Question

a cannonball is fired with a speed of 76 m/s from the top of a cliff. It strikes the plane below with a speed of 89 m/s. if we neglect air friction, how high is the cliff

Answer 1

Assuming the cannonball is fired horizontally, its horizontal velocity stays at a constant 76 m/s. At the point it hits the ground, it has a speed of 89 m/s, so if its vertical velocity at that moment is $v_y$, we have

$89\dfrac{\rm m}{\rm s}=\sqrt{\left(76\dfrac{\rm m}{\rm s}\right)^2+{v_y}^2}\implies{v_y}^2\approx2145\dfrac{\mathrm m^2}{\mathrm s^2}$

(taking the negative square root because we take the downward direction to be negative)

Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are the initial and final velocities, respectively; $a$ is the acceleration; and $\Delta x$ is the change in position of a body. In the cannonball's case, it starts with 0 vertical velocity and is subject to a downward acceleration with magnitude $g=9.80\frac{\rm m}{\mathrm s^2}$. So we have

$2145\dfrac{\mathrm m^2}{\mathrm s^2}-0=-2g\Delta y\implies\Delta y\approx-109.44\,\mathrm m$

(which is negative because we take the cannonball's starting position at the top of the cliff to be the origin) so the cliff is about 109 m high.