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A ball is dropped from a height of 20m and rebounds with a velocity which is 3/4 of the velocity with which it hit the ground....Asked by A GOODSTUDENT
A ball is dropped from a height of 20 meter and rebounded with a velocity which is 3/4 of the velocity with which it hits the ground. What is time interval between the first and second bounces?
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Answered by
plumpycat
Vf ² = Vi ² + 2ad
Vf ² = 0 + 2(9.8)(20)
Vf ² = 392
Vf ~ 19.80 m/s when it lands.
Initial speed coming back up:
(3/4)(19.80) = 14.85 m/s
Equation of motion for second bounce:
d = Vit + (1/2)at ²
d = 14.85t + (1/2)(-9.8)t ²
Substitute d = 0 (to represent the landing height) and solve for t.
Vf ² = 0 + 2(9.8)(20)
Vf ² = 392
Vf ~ 19.80 m/s when it lands.
Initial speed coming back up:
(3/4)(19.80) = 14.85 m/s
Equation of motion for second bounce:
d = Vit + (1/2)at ²
d = 14.85t + (1/2)(-9.8)t ²
Substitute d = 0 (to represent the landing height) and solve for t.
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