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To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If the ball is in contact with the floor for I2.0 ms, (a) what is the magnitude of its average acceleration during that contact and (b) is the average acceleration up or down?

User Haffax
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1 Answer

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Answer:

Part a)


a = 1260.3 m/s^2

Part b)

Direction = upwards

Step-by-step explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as


v_f^2 - v_i^2 = 2 a d


v_1^2 - 0^2 = 2(9.81)(4.0)


v_1 = 8.86 m/s

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as


v_f^2 - v_i^2 = 2 a d


0 - v_2^2 = 2(-9.81)(2.00)


v_2 = 6.26 m/s

Part a)

Average acceleration is given as


a = (v_f - v_i)/(\Delta t)


a = (6.26 - (-8.86))/(12.0 * 10^(-3))


a = 1260.35 m/s^2

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

User Guess
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