Question

A mountain lion jumps to a height of 3.25 m when leaving the ground at an angle of 43.2°. What is its initial speed (in m/s) as it leaves the ground?

Answer 1

Recall that

${v_f}^2={v_i}^2+2a\Delta y$

where $v_i$ and $v_f$ are the lion's initial and final vertical velocities, $a$ is its acceleration, and $\Delta y$ is the vertical displacement.

At its maximum height, the lion has 0 vertical velocity, so we have

$0={v_i}^2-2gy_{\rm max}$

where g is the acceleration due to gravity, 9.80 m/s², and we take the starting position of the lion on the ground to be the origin so that $\Delta y=y_{\rm max}-0=y_{\rm max}$.

Let v denote the initial speed of the jump. Then

$v_i=v\sin(43.2^\circ)=\sqrt{2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(3.25\,\mathrm m)}\implies\boxed{v\approx11.7\dfrac{\rm m}{\rm s}}$