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(\rm m)/(\mathrm s^2)\right)(3.25\,\mathrm m)}\implies\boxed{v\approx11.7(\rm m)/(\rm s)}`