3 Projectiles
Horizontal launches
Let us reflect on what has been done in the last example because it illustrates both the features of projectile motion in the absence of air resistance and a procedure for solving mechanics problems involving projectiles. Instead of the vector method used in Example 1, the relevant projectile motion could have been considered in terms of the separate equations of motion in the horizontal ( $x$ ) and vertical ( $y$ ) directions; these may be written
$\phantom{\rule{2em}{0ex}}\u1e8d=0$ $\phantom{\rule{2em}{0ex}}\xff=g.$
These may be solved separately but the vector method is neater since it shows horizontal and vertical component information at the same time.
The most important features of projectile motion in the absence of air resistance are the constant vertical acceleration and the constant horizontal speed . In projectile problems, the usual procedure is to find the time taken to reach the vertical coordinate position of interest and then to use this time together with the horizontal component of velocity to get the horizontal distance.
Example 2
In an apparatus to demonstrate twodimensional projectile motion, ball bearings are released simultaneously to roll on two identical ramps that are separated vertically. The ramps consist of sloped and horizontal portions of the same length. The ball bearing on the upper ramp becomes a projectile when it reaches the end of the upper ramp while the lower ball bearing rolls along a horizontal channel when it reaches the end of its ramp. The situation is represented in Figure 4.
 What is the speed of each ball bearing at the end of its ramp ( $A$ or $B$ in Figure 4)?
 How does the point at which the upper ball bearing hits the lower channel vary with the height of the upper ramp?
 Where will be the location of the lower ball bearing at the time at which the projectile ball bearing hits the lower channel?
 What assumptions have been made in answering 1., 2. and 3.?
Figure 4 :
Solution

The concepts of kinetic and potential energy and conservation of total energy may be used. At the top of its ramp, each ball bearing will have a potential energy with respect to the bottom of
$mgd$
, where
$m$
is its mass,
$g$
is gravity and
$d$
is the vertical drop from top to bottom of the ramp. Also it will have zero kinetic energy since it is stationary. At the bottom of the ramp, the potential energy will be zero (as long as the thickness of the ramp is ignored) and the kinetic energy will be
$\phantom{\rule{1em}{0ex}}\frac{1}{2}m{u}^{2}$
where
$u$
is the magnitude of the velocity at the bottom of each ramp. So, by conservation of energy,
$\phantom{\rule{2em}{0ex}}mgd=\frac{1}{2}m{u}^{2},$ or $\phantom{\rule{1em}{0ex}}u=\sqrt{2gd}.$
This will be the component of velocity in the direction of the sloping part of the ramp and, in the absence of any losses along the ramp or at the bend where there is a sudden change in momentum, this becomes the horizontal component of velocity at the end of the ramp.
 Since the ramps are identical, both ball bearings will have the same horizontal component of velocity at the ends of their ramps. Suppose that we use coordinates $x$ (horizontal) and $y$ (downward vertical) with the origin at $A$ . The answer to Example 2(b) may be employed without having to start from scratch. This tells us that the coordinates of the projectile ball bearing when $y=h$ are $\left(u\sqrt{\frac{2h}{g}},h\right)$ or, since $u=\sqrt{2gd}$ , the coordinates are $\left(2\sqrt{hd},h\right)$ . Since $d$ is constant, this means that the location of the point at which the projectile ball bearing hits the lower channel varies with the square root of the height of $A$ above $B$ (i.e. with $\sqrt{h}$ ).
 As remarked earlier, the lower ball bearing will have the same horizontal component of velocity ( $u$ ) at $B$ as the projectile ball bearing has at $A$ . Consequently it will travel the same horizontal distance in the same time as the projectile ball bearing. This means that the projectile ball bearing should hit the lower one.
 In 1. the thickness of the ramps has been ignored and the bends in the ramps have been assumed not to introduce any energy losses. In 2. air resistance has been assumed to be negligible. In 1. and 3. rolling friction along the sloping ramp and the horizontal channel has been assumed to be negligible. In fact the effects due to the bends in the ramps will mean that the calculation of horizontal velocity at the end of the ramp is not accurate. However, it can be assumed that identical bends will affect identical ball bearings identically. So the conclusion that the ball from the upper ramp will hit the lower one is still valid (provided rolling friction for the lower ball bearing is comparable to air resistance for the upper ball bearing).
Task!
A crashed car is found on the beach near an unfenced part of sea wall where the top of the wall is 18 m above the beach and the beach is level. The investigating police officer finds that the marks in the beach resulting from the car’s impact with the beach begin at 8 m from the wall and that the vehicle appears to have been travelling at rightangles to the wall. Estimate how fast the vehicle must have been travelling when it went over the wall.
Use $y$ measured downwards as the vertical coordinate. The vector equation of motion is
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{a}=g\underset{\u0332}{j}.$
Integrating once gives
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{v}=gt\underset{\u0332}{j}+\underset{\u0332}{c}.$
The car’s initial vertical velocity component may be assumed to be zero. If the initial horizontal component is represented by $u\underset{\u0332}{i}$ , then $c=u\underset{\u0332}{i}$ and
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{v}=u\underset{\u0332}{i}+gt\underset{\u0332}{j}.$
Integrating again to get position as a function of time,
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}=ut\underset{\u0332}{i}+\frac{1}{2}g{t}^{2}\underset{\u0332}{j}+\underset{\u0332}{d}.$
In accordance with the initial condition that the vertical position is measured from the top of the sea wall, $\underset{\u0332}{d}=\underset{\u0332}{0}$ and
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}=ut\underset{\u0332}{i}+\frac{1}{2}g{t}^{2}\underset{\u0332}{j}.$
Now consider vertical motion only . When $y=18.0$ ,
$\phantom{\rule{2em}{0ex}}t=\sqrt{\frac{2\times 18}{9.81}}=1.916.$
So the car is predicted to hit the beach after 1.916 s. Next consider horizontal motion . During 1.916 s, in the absence of air resistance, the car is predicted to move a horizontal distance of $u\times 1.916$ . This distance is given as 8 m. So
$\phantom{\rule{2em}{0ex}}8=1.916u$
or
$\phantom{\rule{2em}{0ex}}u=4.175.$
So the car is estimated to have left the sea wall at a speed of just over $4\text{m}{\text{s}}^{1}$ (about 15 kph).
There are several complications that may arise when studying and modelling projectile motion. Launch at some angle other than horizontal is the main consideration in the remainder of this Section. For a given launch speed it is possible to find more than one trajectory that can pass through the same target location. Another complication results from launch at a location that is not the origin of the coordinate system used for modelling the motion.
3.1 Angled launches
Vector equations may be used to obtain the position and velocity of a projectile as a function of time if the object is not launched horizontally but with some arbitrary velocity. We shall start by modelling an angled launch from and to a horizontal ground plane. Again it is sensible to use the launch point as the origin of the coordinate system employed. Ignoring air resistance, we shall find expressions for the velocity and position vectors at time $t$ of an object that is launched from ground level ( $\underset{\u0332}{r}$ = $\underset{\u0332}{0}$ ) at velocity $\underset{\u0332}{u}$ with direction $\theta $ above the horizontal. Subsequently we shall find an expression for the time at which the object will hit the ground and the horizontal distance it will have travelled in this time. Finally we will find the coordinates of the highest point on its trajectory (see Figure 5), the angle that will give maximum range and an expression for maximum range in terms of the magnitude and angle of the launch velocity.
Figure 5 :
We use an upwardpointing vertical vector in all of the remaining projectile problems in this Workbook. We start from first principles using Newton’s second law in vector form:
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{F}=m\underset{\u0332}{a}.$ (1.4)
This time, since the initial motion is upward, we choose to point the unit vector $\underset{\u0332}{j}$ upward and so the weight of the projectile may be expressed as $\underset{\u0332}{W}=mg\underset{\u0332}{j}$ . Ignoring air resistance, the weight is the only force on the projectile, so
$\phantom{\rule{2em}{0ex}}m\underset{\u0332}{a}=mg\underset{\u0332}{j}.$ (1.5)
Note the minus sign which is a result of the choice of direction for $\underset{\u0332}{j}$ . After dividing through by $m$ , we obtain the vector equation for the acceleration due to gravity: $\phantom{\rule{1em}{0ex}}\underset{\u0332}{a}=g\underset{\u0332}{j}.$
Recall that $\underset{\u0332}{a}=\frac{d\underset{\u0332}{v}}{dt}$ so, $\phantom{\rule{1em}{0ex}}\frac{d\underset{\u0332}{v}}{dt}=g\underset{\u0332}{j}.$
Integrating this gives
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{v}\left(t\right)=gt\underset{\u0332}{j}+\underset{\u0332}{c}.$
Integrating again gives
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}\left(t\right)=\frac{1}{2}g{t}^{2}\underset{\u0332}{j}+t\underset{\u0332}{c}+\underset{\u0332}{d}$ (1.6)
Since $\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}\left(0\right)=\underset{\u0332}{0}$ , $\underset{\u0332}{d}=\underset{\u0332}{0}.$
Figure 6 :
The initial velocity may be expressed in vector form. Recall from HELM booklet 9 that the component of a vector along a specific direction is given by the dot product of the vector with the unit vector in the direction of interest. The dot product involves the cosine of the angle between the vectors.
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{v}\left(0\right)=\underset{\u0332}{c}=ucos\theta \underset{\u0332}{i}+usin\theta \underset{\u0332}{j}$ (from Figure 6).
Hence
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{v}\left(t\right)=ucos\theta \underset{\u0332}{i}+\left(usin\theta gt\right)\underset{\u0332}{j}$ (1.7)
and
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}\left(t\right)=\frac{1}{2}g{t}^{2}\underset{\u0332}{j}+t\underset{\u0332}{c}+\underset{\u0332}{d}=utcos\theta \underset{\u0332}{i}+\left(utsin\theta \frac{1}{2}g{t}^{2}\right)\underset{\u0332}{j}.$ (1.8)
The vertical component of the position will be zero at the launch and when the projectile hits the ground. This will be true when
$\phantom{\rule{2em}{0ex}}utsin\theta \frac{1}{2}g{t}^{2}=0$ so $t\left(usin\theta \frac{1}{2}gt\right)=0.$
This gives $t=0$ , as it should, or
$\phantom{\rule{2em}{0ex}}t=\frac{2usin\theta}{g}.$
At this value of $t$ the horizontal position coordinate $utcos\theta $ will give the horizontal range, $R$ , so
$\phantom{\rule{2em}{0ex}}R=u\left(\frac{2usin\theta}{g}\right)cos\theta =\frac{2{u}^{2}sin\theta cos\theta}{g},$
or, since $sin\left(2\theta \right)=2sin\theta cos\theta ,$
$\phantom{\rule{2em}{0ex}}R=\frac{{u}^{2}sin2\theta}{g}.$ (1.9)
From this result it is possible to deduce that the maximum range, ${R}_{\text{max}}$ , of a projectile measured at the same vertical level as its launch occurs when $sin\left(2\theta \right)$ has its maximum value, which is 1, corresponding to $\theta =4{5}^{\∘}$ .
So, the maximum range is given by
$\phantom{\rule{2em}{0ex}}{R}_{max}={u}^{2}\u2215g.$ (1.10)
From (1.8), the height ( $y$ coordinate) at any time is given by
$\phantom{\rule{2em}{0ex}}y\left(t\right)=utsin\theta \frac{1}{2}g{t}^{2}.$
To find the maximum height, we can find the value of $t$ at which $\u1e8f\left(t\right)=0.$ Once we have this value of $t$ we can substitute in the expression for $y$ to find the corresponding value of $y$ . Note that the condition $\u1e8f\left(t\right)=0$ at maximum height is the same as asserting that the vertical component of velocity must be zero at the maximum height. Hence, by differentiating $y\left(t\right)$ above, or from (1.7), it is required that
$\phantom{\rule{2em}{0ex}}usin\theta gt=0$ , which gives $t=\frac{usin\theta}{g}.$
The height for any $t$ is given by $y\left(t\right)=utsin\theta \frac{1}{2}g{t}^{2}$ . After substituting $t=\frac{usin\theta}{g}$ this becomes
$\phantom{\rule{2em}{0ex}}y\left(t\right)=\frac{{u}^{2}{sin}^{2}\theta}{g}\frac{{u}^{2}{sin}^{2}\theta}{2g}=\frac{{u}^{2}{sin}^{2}\theta}{2g}.$
Note that the time at which the projectile reaches its maximum height is exactly half the total time of flight ( $t=\frac{2usin\theta}{g}$ ). In the absence of air resistance, the trajectory of the object will be a parabola with maximum height at its vertex which will occur halfway between the launch and the landing i.e. halfway through its flight. The horizontal coordinate of this point will be
$\phantom{\rule{2em}{0ex}}\frac{{R}_{max}}{2}=\frac{{u}^{2}sin2\theta}{2g}.$
So the coordinates of the maximum height are
$\phantom{\rule{2em}{0ex}}\left(\frac{{u}^{2}sin2\theta}{2g},\phantom{\rule{1em}{0ex}}\frac{{u}^{2}{sin}^{2}\theta}{2g}\right).$ (1.11)
If the trajectory corresponds to maximum range, i.e. $\theta =4{5}^{\∘}$ (which means that ${sin}^{2}\theta =\frac{1}{2}$ ), then the maximum height is $\frac{{u}^{2}}{4g}$ at a horizontal distance of $\frac{{u}^{2}}{2g}$ and the maximum range is $\frac{{u}^{2}}{g}$ . Although several of these results are useful, particularly (1.10) for maximum range, and are worth committing to memory, it is more important to remember the method for deriving them from first principles.
Example 3
During a particular downhill run a skier encounters a short but sharp rise that causes the skier to leave the ground at $25\text{m}{\text{s}}^{1}$ at an angle of $3{0}^{\∘}$ to the horizontal. The ground immediately beyond the rise is flat for 60 m. Beyond this the downhill slope continues. See Figure 7. Ignoring air resistance, will the skier land on the flat ground beyond the rise?
Figure 7 :
Solution
In this Example, from (1.9), $R=625sin\left(60\right)\u2215g=55.7$ m. So the skier will land on the flat part of the slope. The horizontal range achieved will be reduced by air resistance but increased if the skier is able to exploit aerodynamic lift from the skis during flight. In the absence of these effects, an effort to leave the ground at a slightly faster speed would be rewarded with the possibility of landing on the continuation of the downhill slope which may be an advantage in racing since it might reduce the interruption caused by the rise while picking up speed. A speed of $26.5\text{m}{\text{s}}^{1}$ at the same angle would mean that the skier lands beyond the flat ground at a range of 62 m from the rise.
A similar method to that used when considering a projectile launched from ‘ground’ level may be used if the projectile is launched at some height above the chosen origin of coordinates. If air resistance is ignored, the governing vector acceleration is still
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{a}\left(t\right)=g\underset{\u0332}{j}$
or
$\phantom{\rule{2em}{0ex}}\frac{d\underset{\u0332}{v}}{dt}=g\underset{\u0332}{j}.$
Integrating this
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{v}\left(t\right)=gt\underset{\u0332}{j}+\underset{\u0332}{c}.$
Integrating again
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}\left(t\right)=\frac{1}{2}g{t}^{2}\underset{\u0332}{j}+t\underset{\u0332}{c}+\underset{\u0332}{d}.$
This time, instead of being launched from $y=0$ , the projectile is launched from $y=H$ . So $\underset{\u0332}{r}\left(0\right)=H\underset{\u0332}{j}$ , and hence $\underset{\u0332}{d}=H\underset{\u0332}{j}$ . As before, the initial velocity may be expressed in vector form.
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{v}\left(0\right)=\underset{\u0332}{c}=ucos\theta \underset{\u0332}{i}+usin\theta \underset{\u0332}{j}.$
Hence $\underset{\u0332}{v}\left(t\right)=ucos\theta \underset{\u0332}{i}+\left(usin\theta gt\right)\underset{\u0332}{j}$ which is the same as (1.8). But, now
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}\left(t\right)=utcos\theta \underset{\u0332}{i}+\left(utsin\theta \frac{1}{2}g{t}^{2}+H\right)\underset{\u0332}{j}$ (1.12)
which differs from (1.8) by the extra $H$ in the $\underset{\u0332}{j}$ component. Note that this single vector equation for $r\left(t\right)$ may be expressed as two separate equations for $x\left(t\right)$ and $y\left(t\right)$ :
$\phantom{\rule{2em}{0ex}}x\left(t\right)=utcos\theta $ $\phantom{\rule{2em}{0ex}}y\left(t\right)=utsin\theta \frac{1}{2}g{t}^{2}+H.$
Example 4
A stone is thrown upwards at $4{5}^{\∘}$ from a height of 1.5 m above flat ground and lands on the ground at a distance of 30 m from the point of launch. Ignoring air resistance, calculate the speed at launch.
Solution
For the particular case of interest $\theta =4{5}^{\∘}$ and the stone lands at a horizontal distance of 30 m from the point of launch. Using these values in the $x\left(t\right)$ component of (1.12) gives $30=\frac{ut}{\sqrt{2}}$ . So the time for which the stone is in the air is $30\sqrt{2}\u2215u$ . Substitution of this time, at which $y=0$ , into the $y\left(t\right)$ component of (1.12) gives
$\phantom{\rule{2em}{0ex}}0=30g{\left(\frac{30}{u}\right)}^{2}+1.5$ or $u=30\sqrt{\frac{g}{31.5}}=16.739\phantom{\rule{1em}{0ex}}{\text{ms}}^{1}.$
The speed of release is about $17\text{m}{\text{s}}^{1}$ .
3.2 Choosing trajectories
So far most of the projectiles we have considered have been launched without any particular control or target. However there are many instances in sport and recreation where there are clearly defined targets for the projectile motion and the trajectory is controlled through the speed and angle of launch. As we shall discover by considering several examples, it is possible to choose more than one path to achieve a given target. First we model a case in which the choice of angle is important.
Consider a projectile launched at an angle $\theta $ to the horizontal. If we take the origin of coordinates at the launch point, then, according to (1.7), the $x$  and $y$ coordinates of the projectile at time $t$ are
$\phantom{\rule{2em}{0ex}}x\left(t\right)=utcos\theta $ $y\left(t\right)=utsin\theta \frac{1}{2}g{t}^{2}.$
Although the path is characterised parametrically in terms of $t$ by these expressions, if we are given $y$ or $x$ or both instead of $t$ , then it is useful to be able to express $y$ in terms of $x$ . We shall eliminate $t$ , by substituting $t=\frac{x}{ucos\theta}$ (which can be deduced from the first equation) in the second equation to give
$\phantom{\rule{2em}{0ex}}y=xtan\theta \frac{1}{2}g{\left(\frac{x}{ucos\theta}\right)}^{2}.$
By using $\frac{1}{{cos}^{2}\theta}={sec}^{2}\theta =1+{tan}^{2}\theta $ , the equation for $y$ may be written in the form:
$\phantom{\rule{2em}{0ex}}y=xtan\theta \frac{1}{2}g{\left(\frac{x}{u}\right)}^{2}\left(1+{tan}^{2}\theta \right).$ (1.13)
Figure 8 :
For given values of $u,\theta $ and $y$ , Equation (1.13) represents a quadratic in $x$ . For a given speed and angle of launch, a given height is achieved at two different values of $x$ . This is a consequence of the parabolic form of the trajectory (Figure 8). For given values of $u,y$ and $x$ , i.e. given a launch velocity and a target location, Equation (1.13) becomes a quadratic in $tan\theta $ i.e.
$\phantom{\rule{2em}{0ex}}\frac{g}{2}{\left(\frac{x}{u}\right)}^{2}{tan}^{2}\theta xtan\theta +\frac{g}{2}{\left(\frac{x}{u}\right)}^{2}+y=0.$ (1.14)
Recall the condition ${b}^{2}>4ac$ for the quadratic $a{t}^{2}+bt+c=0$ to have real roots. As long as
$\phantom{\rule{2em}{0ex}}{x}^{2}>2g{\left(\frac{x}{u}\right)}^{2}\left(\frac{g}{2}{\left(\frac{x}{u}\right)}^{2}+y\right)$ (1.15)
the quadratic will have two real roots.
Figure 9 :
This means that two different choices of angle of launch will cause the projectile to pass through given coordinates $\left(x,y\right)$ ; this is illustrated in Figure 9. If the projectile is launched from $y=H$ in the chosen coordinate system, then the $x\left(t\right)$ part of Equation (1.13) leads to the substitution $t=\frac{x}{ucos\theta}$ as before, but the equation for $y$ becomes
$\phantom{\rule{2em}{0ex}}y=xtan\theta \frac{1}{2}g{\left(\frac{g}{x}\right)}^{2}\left(1+{tan}^{2}\theta \right)+H$ (1.16)
which differs from (1.13) by the addition of $H$ to the righthand side.
Next we will look at two Examples of projectiles with chosen trajectories. In the first Example the influence of initial speed on the trajectory is important; in the second the influence of angle is important.
Example 5
( choosing speed )
As a result of many years of practice, a university teacher, is skilled at throwing screwed up sheets of paper, containing unsatisfactory attempts at setting examination questions, into a cylindrical waste paper bin. She throws at an angle of $2{0}^{\∘}$ above the horizontal and from 1.5 m above the floor. More often than not, the paper balls land in the bin which is 0.2 m high and has a radius of 0.15 m. The bin is placed so that its nearest edge is 3.0 m away (in a horizontal direction) from the point of launch. Model the paper ball as a projectile. Ignore air resistance and calculate the speed of throw that will result in the paper ball entering the bin at the centre of its open end.
Solution
We can choose the origin at the point of launch, with $x$  and $y$ axes as before (see Figure 10). In this case, we need to use the position vector (Equation (1.11)) and find the condition on the speed for the throw to be on target. In particular, we are given that $\theta =2{0}^{\∘}$ and the location of the bin and need to determine the speed of throw necessary for the paper projectile to pass through the centre of the open end of the bin. The centre of the open end of the bin has the coordinates $\left(3.15,1.3\right)$ with respect to the chosen origin. Note the negative value of the vertical coordinate since the top of the bin is located 1.3 m below the chosen origin.
Figure 10 :
At the centre of the bin, using Equation (1.11) and the horizontal position coordinate, we have
$\phantom{\rule{2em}{0ex}}utcos20=3.15.$ so, $t=\frac{3.15}{ucos20}.$
Also, from (1.16) and the vertical coordinate
$\phantom{\rule{2em}{0ex}}utsin20\frac{1}{2}g{t}^{2}=1.3.$
Using $g=9.81\text{m}{\text{s}}^{2}$ , and the expression for $t$ , gives
$\phantom{\rule{2em}{0ex}}3.15tan\left(20\right)\frac{1}{2}9.81\left({\left(3.15\right)}^{2}\u2215{u}^{2}\right){sec}^{2}20=1.3,$
which means that $\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}u=3.15sec\left(20\right)\sqrt{\frac{0.5\left(9.81\right)}{1.3+3.15tan\left(20\right)}}.$
Hence $u=4.746\text{m}{\text{s}}^{1}$ i.e. the academic throws the screwed up paper at about $4.75\text{m}{\text{s}}^{1}$ .
Clearly the motion of screwed up pieces of paper will depend to a significant extent on air resistance. We shall consider how to model resisted motion later (Section 34.3).
Task!
According to the model developed above, for what range of throwing speeds will the academic be successful in getting the paper ball into the basket?
Assume that the time of flight of the screwed up paper balls and the angle of throw do not change. The permitted variation in throwing speed is determined by the horizontal distance $utcos20$ . The screwed up paper ball will not find the bin if this product is less than 3 m or more than 3.3 m. Using $t=\left(3.15\u22154.75\right)cos20$ with $utcos20$ ranging from 3 to 3.3 gives $4.524\le u\le 4.976$ .
Example 6
( choosing angle )
In the game of Tiddlywinks, small plastic discs or counters (‘tiddlywinks’ or ‘winks’) are caused to spring into the air by exerting sharp downward pressure at their edges with another (usually larger) disc called a squidger. By changing the pressure and overlap of the larger disc it is possible to control the velocity at launch of each wink. The object of the game is to ‘pot’ all of the winks into a cup or cylindrical receptacle before your opponent does. One important skill when ‘potting’ is to be able to clear the edge of the collecting cup with the winks. For a given speed at launch, an experienced or successful player will know how the path changes with angle. Suppose that air resistance can be ignored and that a wink may be modelled as a point object and its spin may be ignored.
Given that the nearest edge of the cup is 0.05 m high,
 Calculate (i) the speed of launch such that the maximum height on the maximum range trajectory is 0.05 m and (ii) the associated maximum range.
 Given a launch speed that is $0.1\text{m}{\text{s}}^{1}$ faster than that calculated in 1. find the angle of launch that is likely to be successful for potting the wink when the centre of the cup is 0.1 m from the point of launch.
Solution

The expression for maximum height (Equation (1.11)) may be used to calculate a corresponding speed of launch.
Hence $\frac{{u}^{2}}{4g}=0.05$ , or $u=\sqrt{0.2g}=1.4$ so the required speed of launch is $1.4\text{m}{\text{s}}^{1}$ .
For this trajectory the maximum height is reached at a horizontal distance of $\frac{{u}^{2}}{2g}=0.1$ m from launch and the maximum range is 0.2 m (see Figure 11).
Figure 11 :

Although the wink, travelling on the trajectory shown in Figure 11 would reach the edge of the cup i.e. a height of 0.05 m at 0.1 m range, it would not necessarily enter the cup. The finite size of the wink might mean that it hits the edge of the cup and falls back. The wink is more likely to enter the cup if it is descending when it encounters the cup.
In this case the launch speed and target coordinates are specified so Equation (1.16) can be brought into play. If we take $x=0.1$ m, $u=1.5\text{m}{\text{s}}^{1}$ , and $y\phantom{\rule{1em}{0ex}}\left(=H\right)=0.05$ m, then it turns out that condition (1.15) is satisfied ( ${x}^{2}=0.01$ and $2g{\left(\frac{x}{u}\right)}^{2}\left(\frac{g}{2}{\left(\frac{x}{u}\right)}^{2}+y\right)=0.006$ ) and there are two values for $\theta $ , which are $41.{5}^{\∘}$ and $71.{8}^{\∘}$ .
The corresponding trajectories $y1\left(x\right)$ and $y2\left(x\right)$ are shown in Figure 12. The smaller angle results in the shallower trajectory (solid line). The larger angle produces the required result (dotted line) that the wink is descending at $x=0.1$ m and hence is more likely to enter the cup. This assumes that the cup is at least 2 cm wide.
Figure 12
Example 7
(choosing the angle again)
An engineering student happens to be a fine shotputter. At a tutorial on projectiles he argues that, because he throws from a height of about 2 m, he needs to launch the shot at an angle other than $4{5}^{\∘}$ to get the greatest range. He claims that when launching at $4{5}^{\∘}$ the furthest he can put the shot is to a horizontal distance of 17 m from the launch.
 Calculate the speed at which he releases the shot at $4{5}^{\∘}$ ignoring air resistance.
 Write down an equation for the trajectory of the shot, assuming that the shot is released always at the maximum speed calculated in (a). Set the vertical coordinate to zero and substitute the constant $L$ for the maximum range at the height of launch to obtain a quadratic for the horizontal range $R$ .
 Hence, by differentiating the resulting equation with respect to $R$ , calculate the optimum angle of launch and the maximum range.
Solution

For the particular case of interest
$\theta =4{5}^{\∘}$
and the shot lands at a horizontal distance of 17 m from the point of launch. Using these values, (1.12) gives
$\phantom{\rule{2em}{0ex}}17=utcos\left(4{5}^{\∘}\right)=\frac{ut}{\sqrt{2}}$
So the time for which the shot is in the air, i.e. before it lands, is $17\sqrt{2}\u2215u$ . Substitution of this time (at which $y=0$ ) into the $y$ part of (1.12) gives
$\phantom{\rule{2em}{0ex}}0=17g{\left(\frac{17}{u}\right)}^{2}+2$ or $u=17\sqrt{\frac{g}{19}}=12.2\phantom{\rule{1em}{0ex}}{\text{ms}}^{1}.$
The speed of release is about $12\text{m}{\text{s}}^{1}$ . According to the shotputter this is more or less his maximum speed of release.

The general equation for the trajectory of the shot is (1.16)
$\phantom{\rule{2em}{0ex}}y=xtan\theta \frac{1}{2}g{\left(\frac{x}{u}\right)}^{2}\left(1+{tan}^{2}\theta \right)+H.$
Given that the maximum speed of release and optimum angle of launch are employed, the shot should land at the maximum range, $R$ . From the general Equation (1.16), with $x=R$ and $y=0$ , we have
$\phantom{\rule{2em}{0ex}}0=Rtan\theta \frac{1}{2}g{\left(\frac{R}{u}\right)}^{2}\left(1+{tan}^{2}\theta \right)+H.$
Since $\frac{{u}^{2}}{g}$ does not depend on either $R$ or $\theta $ , we replace it by a constant $L$ , and rearrange the equation into the usual form for a quadratic in $R$ :
$\phantom{\rule{2em}{0ex}}0=\frac{{R}^{2}}{2L}\left(1+{tan}^{2}\theta \right)+Rtan\theta +H.$

The optimum angle of launch is found by obtaining an expression for
$R$
and setting
$\frac{dR}{d\theta}$
equal to zero. As you can imagine, the expression for
$R$
resulting from solving this quadratic is rather complicated and nasty to differentiate. An alternative approach is called
implicit differentiation
. (See
HELM booklet
11.7). We work through the equation as it stands differentiating term by term with respect to
$\theta $
and making use of the relationship
$\phantom{\rule{2em}{0ex}}\frac{df\left(R\right)}{d\theta}=\frac{df\left(R\right)}{dR}\times \frac{dR}{d\theta}.$
(For example, $\frac{d\left({R}^{2}\right)}{d\theta}=2R\frac{dR}{d\theta}$ ). Hence implicit differentiation gives
$\phantom{\rule{2em}{0ex}}0=\frac{1}{2L}\left(2R\frac{dR}{d\theta}\right)\left(1+{tan}^{2}\theta \right)\frac{{R}^{2}}{2L}\left(2tan\theta {sec}^{2}\theta \right)+\frac{dR}{d\theta}tan\theta +R{sec}^{2}\theta .$
At the maximum range, $\frac{dR}{d\theta}=0$ :
$\phantom{\rule{2em}{0ex}}0=\frac{{R}^{2}}{2L}\left(2tan\theta {sec}^{2}\theta \right)+R{sec}^{2}\theta $ so $R{sec}^{2}\theta \left(\frac{R}{L}tan\theta +1\right)=0.$
Since ${sec}^{2}\theta $ cannot be zero and the option of $R=0$ is not very interesting, it is possible to conclude that the relationship between the optimum angle of launch and the maximum range is given by
$\phantom{\rule{2em}{0ex}}tan\theta =\frac{L}{R}.$
This result may be substituted back into the quadratic for $R$ to give
$\phantom{\rule{2em}{0ex}}0=\frac{{R}^{2}}{2L}\left(1+\frac{{L}^{2}}{{R}^{2}}\right)+L+H$ or $0=\frac{{R}^{2}}{2L}\frac{L}{2}+L+H.$
Multiplying throughout by $2L$ gives
$\phantom{\rule{2em}{0ex}}{R}^{2}={L}^{2}+LH$ i.e. $R=\sqrt{\left({L}^{2}+2LH\right)}.$
A consequence of this result is that $R>L$ . Bearing in mind that $L={u}^{2}\u2215g$ is the maximum range at the height of the launch (or for a launch at ground level), this means that the maximum range from the elevated launch to ground level is greater than the maximum range in the plane at the height of the launch. Substituting the result for $R$ in the result for $tan\theta $ gives
$\phantom{\rule{2em}{0ex}}tan\theta =\frac{L}{\sqrt{{L}^{2}+2HL}}=\frac{1}{\sqrt{1+\frac{2H}{L}}}.$
For $H>0$ , this implies that $tan\theta <1$ , which in turn implies an optimum angle of launch $<4{5}^{\∘}$ and that the shot putter’s assertion is justified. When a projectile is launched at some angle from some point above ground level to land on the ground then the optimum angle of launch is less than $4{5}^{\∘}$ . Specifically, if $u=12.2\text{m}{\text{s}}^{1}$ and $H=2$ m, then $L=15.2$ m, $R=17.06$ m and $tan\theta =0.9\phantom{\rule{1em}{0ex}}\left(\theta =41.{7}^{\∘}\right)$ .
The $4{5}^{\∘}$ launch trajectory and optimum angle launch trajectory are shown in Figure 13 together with the maximum range trajectory for a ground level launch. A close up of the ends of the first two trajectories is shown in Figure 14. The shotputter can increase the length of his putt only by a few centimetres if he putts at the optimum angle of launch rather than $4{5}^{\∘}$ . However these could be a vital few centimetres in a tight competition!
Figure 13 :
Figure 14 :
Task!
A fairground stall known as a ‘coconut shy’ consists of an array of coconuts placed on stands. The objective is to win a coconut by knocking it off its stand with a wooden ball. A local youth has learned that if he throws a wooden ball as fast as he can at $1{0}^{\∘}$ above the horizontal he is able to hit the nearest coconut more or less dead centre and knock it down almost every time. The nearest coconut stand is located 4 m from the throwing position with its top at the same height as the balls are thrown. The coconuts are 0.1 m long.

Calculate how fast the youth is able to throw if air resistance is ignored:
Choose an origin of coordinates at the point of launch of the wooden balls, with the $y$ coordinate in the vertical upwards direction and the $x$ coordinate along the horizontal towards the coconut.
If air resistance is ignored, the trajectory of the balls may be modelled by the Equation (1.13)
$\phantom{\rule{2em}{0ex}}y=xtan\theta \frac{1}{2}g{\left(\frac{x}{u}\right)}^{2}\left(1+{tan}^{2}\theta \right).$
If the balls hit the coconut dead centre, then their trajectory must pass through the coordinates (4, 0.05) (see diagram above). Hence
$\phantom{\rule{2em}{0ex}}y=xtan\theta \frac{1}{2}g{\left(\frac{x}{u}\right)}^{2}\left(1+{tan}^{2}\theta \right)$
so $\phantom{\rule{2em}{0ex}}{\left(\frac{4}{u}\right)}^{2}=\frac{2}{g}\frac{\left(4tan100.05\right)}{1+{tan}^{2}10}$ or $\phantom{\rule{2em}{0ex}}u=4\sqrt{\frac{g}{2}\frac{1+{tan}^{2}10}{\left(4tan100.05\right)}}=11.1.$
So the youth is able to throw at $11.1\text{m}{\text{s}}^{1}$ .

Calculate how much further the operator of the fairground stall should move the cocunuts from the throwing line to prevent the youth hitting the coconut so easily:
The youth will fail if the nearest coconut is moved sufficiently far away so that the trajectory considered in part 1. passes beneath the bottom of the coconut on top of the stand. If $x$ is the horizontal range corresponding to a $y$ coordinate of 0, then, using the expression in Equation (1.9) on page 13:
$\phantom{\rule{2em}{0ex}}x=\frac{{u}^{2}sin2\theta}{g}.$ In this case, $\phantom{\rule{2em}{0ex}}x=\frac{{\left(11.1\right)}^{2}sin20}{9.81}=4.296.$
So the nearest coconut stand should be moved another 0.296 m from the throwing line. This has assumed that the youth will favour as ‘flat’ a trajectory as possible. The youth could choose to throw at a greater angle to increase the range. For example throwing at an angle of $4{5}^{\∘}$ would result in a range of 12.6 m. However the steeper the angle of launch, the greater will be the angle to the horizontal at which the ball arrives at the coconut. A large angle would not be as efficient as a small one for dislodging it.
Task!
Basketball players are able to gain three points for longrange shooting. The shot must be made from outside a certain radius from the basket. A skilled player makes a jump shot rather than standing on the ground to shoot. He leaps so that he is able to project the ball at a slower, i.e. more controllable, speed and from the same height as the basket, which is 3 m above the ground. Assume that the ball would be released from a height of 2 m when the player is standing on the ground and that air resistance can be ignored.

Calculate the speed of release during a jump shot made at a horizontal distance of 12 m from the basket at maximum range for that speed of release:
If the maximum range is 12 m, then, since the jump shot is made at the same height as the basket, ${u}^{2}\u2215g=12$ , i.e. $u=10.85$ , so the speed of release is $10.85\text{m}{\text{s}}^{1}$ .

Calculate the preferred angle of launch that would hit the basket if the shot were to be made when the player is standing at the same point and shoots at
$12\text{m}{\text{s}}^{1}$
:
Use may be made of Equation (1.15), in the form of a quadratic for $tan\left(\theta \right)$ , where $\theta $ is the launch angle, with $y=3,x=12,H=2$ . The largest root corresponds to the preferred form of trajectory for passing into the basket (see diagram below) and gives $tan\left(\theta \right)=1.764$ or $\theta =60.{5}^{\∘}$ .