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netineya [11]
2 years ago
12

A ball of plasticine is released from rest at height of 2.2 m above the ground. After touching the ground, the plasticine ball c

omes to rest in 96 ms. Determine the average acceleration (magnitude and direction) of the plasticine ball while coming to rest
Physics
1 answer:
Anna35 [415]2 years ago
4 0

The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²

The direction of the acceleration of the ball is downwards

The given parameters

initial velocity of the ball, u = 0

height above the ground, h = 2.2 m

time of motion of the ball, t = 96 ms = 0.096 s

The magnitude of the acceleration of the ball while coming to rest is calculated as;

let the downwards direction of the acceleration be positive

h = ut + 0.5 at^2\\\\h = 0 + 0.5at^2\\\\h = 0.5 at^2\\\\a = \frac{h}{0.5t^2} \\\\a = \frac{2.2}{0.5 \times 0.096^2} \\\\a = 477.43 \ m/s^2

The direction of the acceleration of the ball is downwards

Learn more here: brainly.com/question/15407740

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