Question

A ball is dropped on to a floor from a height of 10 m. It rebounds to a height of 2.5 m . If the ball is in contact with the floor for 00.01 sec. What is the average acceleration during contact.
(A) 1400 m/s^2
(B) 2100 m/s^2
(C) 700 m/s^2
(D) 2800 m/s^2

Answer 1

when it is dropped from 10m,
Initial height = 10m
initial velocity = 0
velocity just before hotting ground = √2gh = √2*9.8*10 = 14.07 m/s (downward)

after rebound,
maximum height reached = 2.5m
final velocity at top = 0
initial velocity(just after rebound) = √2gh = √2*9.8*2.5 = √49 = 7 m/s (upward)

assuming downward as positive direction
So velocity just before hitting ground = +14.07 m/s 
velocity just after hitting ground = -7 m/s 
change in velocity = +14.07 - (-7) = 21.07 m/s
time = 0.01s
acceleration = change in velocity/time = 21.07/0.01 = 2107 m/s² ~ 2100m/s²
Answer is B

Answer 2

Answer:

when it is dropped from 10m,

Initial height = 10m

initial velocity = 0

velocity just before hotting ground = √2gh = √2*9.8*10 = 14.07 m/s (downward)

after rebound,

maximum height reached = 2.5m

final velocity at top = 0

initial velocity(just after rebound) = √2gh = √2*9.8*2.5 = √49 = 7 m/s (upward)

assuming downward as positive direction

So velocity just before hitting ground = +14.07 m/s 

velocity just after hitting ground = -7 m/s 

change in velocity = +14.07 - (-7) = 21.07 m/s

time = 0.01s

acceleration = change in velocity/time = 21.07/0.01 = 2107 m/s² ~ 2100m/s²

Answer is B