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:
[tex]s=u_y t+\frac{1}{2}at^2[/tex]
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,
[tex]u_y = 0[/tex] (initial vertical velocity is zero)
s = 20 m
[tex]a=g=9.8 m/s^2[/tex] (acceleration of gravity)
And solving for t, we find the time of flight:
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s[/tex]
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:
[tex]v_x = 7 m/s[/tex]
Therefore the distance covered after time t is given by
[tex]d=v_x t[/tex]
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:
[tex]d=(7)(2.02)=14.1 m[/tex]
C) 21.0 m/s at [tex]70.5^{\circ}[/tex] 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:
[tex]v_x = 7 m/s[/tex]
Instead, the vertical velocity is given by
[tex]v_y = u_y +at[/tex]
And substituting t = 2.02 s, we find the vertical velocity at the moment of impact:
[tex]v_y = 0+(9.8)(2.02)=19.8 m/s[/tex]
So, the magnitude of the final velocity is
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{7^2+(19.8)^2}=21.0 m/s[/tex]
And the angle is given by
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{19.8}{7})=70.5^{\circ}[/tex]
below the horizontal
