Question

A fuzzy bunny sees a butterfly and chases it right off of a 20 m high cliff at 7m/s

Answer 1

A) 2.02 s

To find the time it takes for the bunny to reach the rocks below the cliff, we just need to analyze its vertical motion. This motion is a free fall motion, which is a uniformly accelerated motion. So we can use the suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where

s is the vertical displacement

u is the initial vertical velocity

a is the acceleration

t is the time

For the bunny here, choosing downward as positive direction,

$u_y = 0$ (initial vertical velocity is zero)

s = 20 m

$a=g=9.8 m/s^2$ (acceleration of gravity)

And solving for t, we find the time of flight:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s$

B) 14.1 m

For this part, we need to consider the horizontal motion of the bunny.

The horizontal motion of the bunny is a uniform motion with constant velocity, which is the initial velocity of the bunny:

$v_x = 7 m/s$

Therefore the distance covered after time t is given by

$d=v_x t$

And substituting the time at which the bunny hits the ground,

t = 2.02 s

We find how far the bunny went from the cliff:

$d=(7)(2.02)=14.1 m$

C) 21.0 m/s at $70.5^{\circ}$ below the horizontal

The horizontal component of the velocity is constant during the entire motion, so it will still be the same when the bunny hits the ground:

$v_x = 7 m/s$

Instead, the vertical velocity is given by

$v_y = u_y +at$

And substituting t = 2.02 s, we find the vertical velocity at the moment of impact:

$v_y = 0+(9.8)(2.02)=19.8 m/s$

So, the magnitude of the final velocity is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{7^2+(19.8)^2}=21.0 m/s$

And the angle is given by

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{19.8}{7})=70.5^{\circ}$

below the horizontal