2 Projectiles: an introduction
2.1 Vertical motion under gravity
Consider a marble which is thrown horizontally off the Clifton Suspension Bridge at a speed of $10\text{m}{\text{s}}^{1}$ and falls into the River Avon. We wish to find the location at which it will splash into the river. We assume that the only force acting on the marble is the force of gravity and that this force is constant. The marble is regarded as a projectile i.e. a point object which has mass but does not spin or rotate. Another assumption made is that the Earth is locally flat. Since the initial vertical speed is zero, application of the distancetime equation ( $s=ut+\frac{1}{2}a{t}^{2}$ ) to the vertical motion gives
$\phantom{\rule{2em}{0ex}}y\left(t\right)=\frac{1}{2}g{t}^{2}$ (1.1)
where $y$ is measured downwards from the bridge and $g$ is the acceleration due to gravity.
The position vector of an object falling freely in the vertical ( $\underset{\u0332}{j}$ ) direction with zero initial velocity and no air resistance may be expressed as a position vector $\underset{\u0332}{r}\left(t\right)$ which is a variable vector depending on the (scalar) variable $t$ representing the time where
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}\left(t\right)=y\left(t\right)\underset{\u0332}{j}=\frac{1}{2}g{t}^{2}\underset{\u0332}{j},$ illustrated in the diagram below.
For motion in a straight line there is no particular reason for introducing vectors. However, timedependent vectors may be used to describe more complicated motion  for example that along curved paths. By introducing the horizontal unit vector $\underset{\u0332}{i}$ in addition to the vertical unit vector $\underset{\u0332}{j}$ , a position vector in two dimensions may be written
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}\left(t\right)=x\left(t\right)\underset{\u0332}{i}+y\left(t\right)\underset{\u0332}{j}.$
For an object falling vertically , $x\left(t\right)=0$ because $x$ does not change with time. Suppose, however, that the object were to have been launched horizontally at speed $u$ . Then, if air resistance is ignored and there are no other forces acting in the horizontal direction, the horizontal acceleration is zero and the horizontal speed of the object should remain constant. This means that the horizontal coordinate is given by $x\left(t\right)=ut$ and, using the earlier result for $y\left(t\right)$ , the vector function describing the position at time $t$ of an object thrown horizontally from some point, which is taken as the origin of coordinates, is given by
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}\left(t\right)=ut\underset{\u0332}{i}+\frac{1}{2}g{t}^{2}\underset{\u0332}{j}.$ (1.2)
The coordinate system and the vectors corresponding to such a situation are shown in Figure 1.
Figure 1 :
The information in Equation (1.2) is sufficient to determine the object’s position graphically at any time $t$ , since it gives both $x\left(t\right)$ and $y\left(t\right)$ and hence it is possible to plot $y\left(t\right)$ against $x\left(t\right)$ for various values of $t$ . An example calculation of the path during the first 3.5 s of the descent of an object thrown horizontally at $10\text{m}{\text{s}}^{1}$ with $g=9.81\text{m}{\text{s}}^{2}$ is shown in Figure 2. In this graph values of $y\left(t\right)$ increase downward so that the curve corresponds to the downward path of the object. The technical name used to describe such a path is the trajectory .
Figure 2
Given the vector components of the timedependent velocity, it is possible to calculate its magnitude and direction at any time. The magnitude is given by the square root of the sum of the squares of the components. Hence, from the last example, the magnitude of the position vector is given by
$\phantom{\rule{2em}{0ex}}\left\underset{\u0332}{r}\right={\left({u}^{2}{t}^{2}+\frac{1}{4}{g}^{2}{t}^{4}\right)}^{1\u22152}.$
The angle of the position vector measured clockwise from the $x$ direction is given by
$\phantom{\rule{2em}{0ex}}cos\mathrm{\ϕ}=ut\u2215\left\underset{\u0332}{r}\right\phantom{\rule{2em}{0ex}}sin\mathrm{\ϕ}=\frac{1}{2}g{t}^{2}\u2215\left\underset{\u0332}{r}\right$
so
$\phantom{\rule{2em}{0ex}}tan\mathrm{\ϕ}=\frac{1}{2}gt\u2215u.$
Note that the angle is zero when $t$ is zero and increases with $t$ (as might be expected). Figure 3 shows graphs of $\left\underset{\u0332}{r}\right$ and $\mathrm{\ϕ}$ for the example and values of $t$ considered in Figure 2.
Figure 3 :
Note that $\mathrm{\ϕ}$ is the angle that the position vector makes with the horizontal and does not denote the direction of motion (i.e. the velocity) of the object. Note also that by introducing another unit vector $\underset{\u0332}{k}$ at rightangles to both $\underset{\u0332}{i}$ and $\underset{\u0332}{j}$ it is possible to consider motion in three dimensions.
Task!
Write down the position vector for a particle moving so that its coordinates are given by
$\phantom{\rule{2em}{0ex}}x=2cos\left(wt\right)\phantom{\rule{2em}{0ex}}y=2sin\left(wt\right)\phantom{\rule{2em}{0ex}}z=1.$
What is the corresponding magnitude of this vector? How would you describe the resulting motion?
The position vector may be written
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}\left(t\right)=2cos\left(\omega t\right)\underset{\u0332}{i}+2sin\left(\omega t\right)\underset{\u0332}{j}+\underset{\u0332}{k}.$
Hence
$\phantom{\rule{2em}{0ex}}\left\underset{\u0332}{r}\left(t\right)\right=\sqrt{4{cos}^{2}\left(\omega t\right)+4{sin}^{2}\left(\omega t\right)+1}=\sqrt{5}.$
Since this is constant, the particle stays at a constant distance from the origin during its motion. When $t=0,\phantom{\rule{1em}{0ex}}\underset{\u0332}{r}\left(0\right)=2\underset{\u0332}{i}+\underset{\u0332}{k}$ .
The object is moving in a circle of radius 2 in the $z=1$ plane (see diagram).
Task!
Show that the vector function
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}\left(t\right)=at\underset{\u0332}{i}+b{t}^{2}\underset{\u0332}{j},$
where $a$ and $b$ are constant scalars, can be represented by a parabola.
By comparing Equation (1.1) with the equation in this Task demonstrate that the trajectory shown in Figure 2 is part of a parabola.
Given $\underset{\u0332}{r}\left(t\right)=at\underset{\u0332}{i}+b{t}^{2}\underset{\u0332}{j}=x\left(t\right)\underset{\u0332}{i}+y\left(t\right)\underset{\u0332}{j}$ we can write $x\left(t\right)=at$ and $y\left(t\right)=b{t}^{2}$ . Using the first of these to obtain $t=\frac{x}{a}$ and substituting for $t$ in the second, we obtain $y=b\frac{{x}^{2}}{{a}^{2}}=c{x}^{2}$ where $c$ is a constant. This has the form of a parabola centred on $\left(0,0\right)$ .
Suppose that we wish to calculate the coordinates at which the marble will splash into the River Avon, given that the water surface is 77 m below the point of launch. Since the horizontal component of velocity is not changing during the fall, we concentrate on the vertical motion. The strategy is to calculate the length of time it takes to drop through the vertical distance between the point of launch and the water surface and then use this time to calculate the horizontal distance moved at constant speed. We use Equation (1.1) to calculate the length of time needed to fall 77 m i.e. the value of $t$ such that
$\phantom{\rule{2em}{0ex}}\frac{1}{2}g{t}^{2}=77.$
This gives $t=\sqrt{\frac{2\times 77}{9.81}}=3.96$ s. During this time the marble will have moved a horizontal distance $ut$ . So if $u=10\text{m}{\text{s}}^{1}$ , the horizontal distance moved is 39.6 m and the coordinates of the splash down are (39.6, 77.0).
The question arises of how to deal with more general problems of a similar nature but starting from first principles. This question leads to a fuller consideration of vector representations of motion.
2.2 Velocity and acceleration vectors
The first derivative of timedependent position vectors may be identified as the velocity vector and the second derivative as the acceleration vector. So, for the example of a stone falling from rest under gravity without air resistance, given that the velocity vector is the first derivative of the position vector,
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{v}\left(t\right)=\frac{d}{dt}\underset{\u0332}{r}\left(t\right)=\frac{d}{dt}\left(\frac{1}{2}g{t}^{2}\underset{\u0332}{j}\right)=gt\underset{\u0332}{j}$ (since $\underset{\u0332}{j}$ does not vary with $t$ ).
Similarly, the acceleration vector is the second derivative of the position vector, which will be the same as the first derivative of the velocity vector, so
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{a}\left(t\right)=\frac{d}{dt}\underset{\u0332}{v}\left(t\right)=\frac{{d}^{2}}{d{t}^{2}}\underset{\u0332}{r}\left(t\right)=g\underset{\u0332}{j}.$
Note that this is an expected result (the acceleration is that due to gravity).
In two dimensions
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{v}\left(t\right)=\frac{dx}{dt}\underset{\u0332}{i}+\frac{dy}{dt}\underset{\u0332}{j}$
and
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{a}\left(t\right)=\frac{{d}^{2}x}{d{t}^{2}}\underset{\u0332}{i}+\frac{{d}^{2}y}{d{t}^{2}}\underset{\u0332}{j}.$
For the marble thrown horizontally at velocity $u$ from the bridge
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{v}\left(t\right)=u\underset{\u0332}{i}+gt\underset{\u0332}{j}$
and
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{a}\left(t\right)=g\underset{\u0332}{j}.$
Note that the horizontal and vertical parts of the velocity (or acceleration) are called the horizontal and vertical components respectively. For the marble thrown horizontally at speed $u$ from the bridge the horizontal component of velocity at any time $t$ is $u$ and the vertical component of velocity at any time $t$ is $gt$ .
Since each component of the vector is differentiated separately, the integral of the acceleration vector may be identified with a velocity vector and the integral of the velocity vector may be identified with a position vector. These give the same expressions as those that we started with apart from arbitrary constants. Note that when integrating vector expressions the arbitrary constant is a constant vector .
Example 1
 Use integration and the variables and vectors identified in Figure 1 to derive vector expressions for the velocity and position of an object thrown horizontally from a bridge at speed $u$ ignoring air resistance.
 Find the object’s coordinates after it has dropped a distance $h$ .
Solution

The acceleration has only a vertical component i.e. the acceleration due to gravity.
$\underset{\u0332}{a}\left(t\right)=g\underset{\u0332}{j}.$
Integrating once gives $\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\underset{\u0332}{v}\left(t\right)=gt\underset{\u0332}{j}+\underset{\u0332}{c}$ where $\underset{\u0332}{c}$ is a constant vector.
The initial velocity has only a horizontal component, so $\phantom{\rule{1em}{0ex}}\underset{\u0332}{v}\left(0\right)=\underset{\u0332}{c}=u\underset{\u0332}{i}$ and $\phantom{\rule{1em}{0ex}}\underset{\u0332}{v}\left(t\right)=u\underset{\u0332}{i}+gt\underset{\u0332}{j}.$
Integrating again $\phantom{\rule{1em}{0ex}}\underset{\u0332}{r}\left(t\right)=ut\underset{\u0332}{i}+\frac{1}{2}g{t}^{2}\underset{\u0332}{j}+\underset{\u0332}{d}$ where $\underset{\u0332}{d}$ is another constant vector.
Since $\underset{\u0332}{r}\left(0\right)=\underset{\u0332}{0}$ , then $\underset{\u0332}{d}=\underset{\u0332}{0}$ (the zero vector), so
$\phantom{\rule{2em}{0ex}}\underset{\u0332}{r}\left(t\right)=ut\underset{\u0332}{i}+\frac{1}{2}g{t}^{2}\underset{\u0332}{j}$
which is the result obtained previously as Equation (1.2) by considering the horizontal and vertical components of motion separately.

The position coordinates at any time
$t$
are
$\phantom{\rule{1em}{0ex}}\left(ut,\frac{1}{2}g{t}^{2}\right).$
When $y\left(t\right)=h$ , then $\frac{1}{2}g{t}^{2}=h$ , or
$\phantom{\rule{2em}{0ex}}t=\sqrt{\frac{2h}{g}}.$ (1.3)
At this value of $t$ , $\phantom{\rule{1em}{0ex}}x\left(t\right)=ut=u\sqrt{\frac{2h}{g}}.$ So the coordinates when $y\left(t\right)=h$ are $\left(u\sqrt{\frac{2h}{g}},h\right)$ .
In this Workbook you will only meet straightforward examples of vector integration where the integral of the vector is obtained by integrating its components. More complicated vector integrals called line integrals are introduced in HELM booklet 29.