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

find the velocities just before impact and just after impact...
acceleration is (change in velocity/time)
write answer na
Someone plzz :( answer
thanks in advance


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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
3 5 3
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she is in 11th
12th, one class up ;)