4 Energy and projectile motion

In this Section we demonstrate that, in the absence of air resistance, energy is conserved during the flight of a projectile. Consider first the launch of an object vertically with speed $u$ . In the absence of air resistance, the height reached by the object is given by the result obtained in Equation (1.9) on page 13 i.e. $\frac{u^2{sin}^2\theta }{2g}$ with $\theta =90^{\circ }$ , so the height is $\frac{u^2}{2g}$ . This result was obtained in Equation (1.10) by considering time of flight. However it can be obtained also by considering the energy. If the highest point reached by the object is $h$ above the point of launch, then with respect to the level of launch, the potential energy of the object at the highest point is $mgh$ . Since the vertical velocity is zero at this point then $mgh$ represents the total energy also. At the launch, the potential energy is zero and the total energy is given by the kinetic energy $\frac{1}{2}m{u}^2$ . Hence, according to the conservation of energy

$\frac{1}{2}m{u}^2=mgh$ or $h=\frac{u^2}{2g}$

as required. Now we will repeat this analysis for the more general case of an angled launch and for any point $\left(x,y\right)$ along the trajectory. Let us use the general results for the position vector and velocity vector expressed in Equations (1.4) and (1.5) on page 12. In the absence of air resistance, the horizontal component of the projectile velocity $\left(v\right)$ is constant. If the height of the launch is taken as the reference level, then the potential energy at any time $t$ and height $y$ is given by

$mgy=mg\left(utsin\theta -\frac{1}{2}g{t}^2\right).$

The kinetic energy is given by

So

$\frac{1}{2}m{\left|v\right|}^2+mgx=\frac{1}{2}m{u}^2.$

But the initial kinetic energy, which is the initial total energy also, is given by $\frac{1}{2}m{u}^2$ . Consequently we have shown that energy is conserved along a projectile trajectory.