Worked example 3.3: Cannon shot

Question: A cannon placed on a 50m high cliff fires a cannonball over the edge of the cliff at $v=200 {\rm m}/{\rm s}$ making an angle of $\theta = 30^\circ$ to the horizontal. How long is the cannonball in the air? Neglect air resistance.
 
Answer: In order to answer this question we need only consider the cannonball's vertical motion. At $t=0$ (i.e., the time of firing) the cannonball's height off the ground is $z_0=50$m and its velocity component in the vertical direction is $v_0 = v \sin\theta = 200\times\sin 30^\circ = 100 {\rm m}/{\rm s}$. Moreover, the cannonball is accelerating vertically downwards at $g=9.91 {\rm m}/{\rm s}^2$. The equation of vertical motion of the cannonball is written

$$z = z_0 + v_0 t - \frac{1}{2} g t^2,$$

where $z$ is the cannonball's height off the ground at time $t$. The time of flight of the cannonball corresponds to the time $t$ at which $z=0$. In other words, the time of flight is the solution of the quadratic equation

$$0 = z_0 + v_0 t - \frac{1}{2} g t^2.$$

Hence,

$$t = \frac{v_0 +\sqrt{v_0^{ 2}+2 g z_0}}{g} = 20.88 {\rm s}.$$

Here, we have neglected the unphysical negative root of our quadratic equation.