An apple drops from the tree and falls freely. The apple is originally at rest a height H0 above the ground. The ground is covered with a thick lawn, which is made of blades of grass of height h. When the apple enters the grass, it slows down at a constant rate so that its speed is 0 when it reaches ground level.

A)Find the speed of the apple just before it enters the grass

B) Find the acceleration of the apple while it is in the grass

Explanation:

The distance travelled in the free fall is H - h  

Since the apple originally started from rest we can use v^2 =  u^2 + 2 x g x s  where v is the final velocity, g the accln due to gravity and s the distance travelled and u is the initial velocity = 0

So the velocity just before it enters the grass is sq rt [2 x g x (H - h)]

Once in the grass, it slows down at a constant rate which means that the acceleration (a) during this period is constant.

So once again using the same formula we have v = O and u = sq rt[2 x g x (H-h)]

so since v^2 = u^2 + 2 x a x s then  

O^2 = 2 x g x (H-h) + 2 x a x h

{O^2 - 2 x g x (H - h)}/(2 x h) = a

the answer is a. hope this .