Question

Napoleon’s cannon, having a muzzle velocity of 126 m/s, is fired at a 0 degree angle off a cliff. The ball lands 739 m away from the cannon. How far above ground level is the cannon?

Answer 1

Let us write equations of motion of cannon:

$y(t) = H - \frac{g t^2}{2}$, $x(t) = v_0 t$ , where $H$ is the height of the cliff, that we need to find, $v_0 = 126 \frac{m}{s}$ is the muzzle velocity, $S = 739 m$ is the horizontal distance, that the ball covered.

When the ball reaches the ground in time $t_1$, $y(t_1) = 0$, from where using equation for y:$0 = H - \frac{g t_1^2}{2}$ , hence the height of the cliff is $H = \frac{g t_1^2}{2}$.

Let us find time $t_1$ using equation for x:

At time $t_1$, $x(t_1) = S = v_0 t_1$, from where $t_1 = \frac{S}{v_0}$.

Substituting last expression for $t_1$ into formula for $H$, obtain:

$H = \frac{g t_1^2}{2} = \frac{g}{2} \frac{S^2}{v_0^2} \approx 168.7 m$