# What is the answer of the question?

Given, the height from which the ball is dropped, h1 =10 m,

The height to which the ball rebounds, h2 = 2.5 m

Time of contact of the ball with floor, Δt = 0.01 sec

Let v1 be the velocity of the ball when dropped before striking the floor,

Let v2 be the velocity of ball upwards after striking the floor during rebounding

Change in velocity of ball during contact = v2 -(-v1)

Now,acceleration a = (v2 – (-v1))/Δt ….(i)

When a body is dropped from height h1:

u = 0, v = v1, a = g, S = h1

The equation of motion: v^2 = u^2 + 2as

v1^2 = 0 + 2gh1

v1 = √(2gh1) ….. (ii)

Consider the motion of ball after striking the floor:

u = v2, v = 0, a = -g, S = h2

The equation of motion becomes: 0 = v2^2+2(-g)h2

v2 = √(2gh2) ….(iii)

Substituting (ii) and (iii) in (i), we get,

a = [√(2gh2) + √(2gh1)]/Δt

a = [√(2×9.8×2.5) + √(2×9.8×10)]/0.01

a = 2100 m/s^2

Hence, average acceleration is 2100 m/s^2