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Grade: 11

                        

An object is dropped from the top of a building 5 metres and rebounds to a height of 3.2 metre if it is in contact with the floor for 0.036 second, what is its average acceleration during this period ? (g=10m per sec sq)

2 years ago

Answers : (1)

Arun
24742 Points
							

While falling down, initial velocity = 0, final velocity = v, acceleration due to gravity = 9.8 m/s2, height = 5 m

The velocity of the ball with which it hits the ground can be found using,

v2 = u2 + 2as

=> v2 = 0 + 2×9.8×5

=> v = 9.9 m/s (downward)

While it moves up, initial velocity = u, final velocity = 0, acceleration due to gravity = 9.8 m/s2, height = 3.2 m

The velocity of the ball with which it leaves the ground can be found using,

v2 = u2 + 2as

=> 0 = u2 - 2×9.8×3.2 [this time the velocity and acceleration are oppositely directed so we have the negative sign]

=> v = 7.91 m/s (upward)

So, Change in velocity = 7.91 m/s (upward) – 9.9 m/s (downward)

=> Δv = 7.91 m/s + 9.9 m/s  [both in upward direction]

=> Δv = 17.81 m/s

So, acceleration is, a = Δv/Δt = 17.81/0.036 = 494.7 m/s2

2 years ago
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