A ball is dropped from a height of 2.50 m above the floor. a) Find the speed v with which it reaches the floor. b) The ball now rebounds. The speed of the ball is decreased to 3v/4 due to this collision. How high will the ball rise?
Answers
so a) 2.50by t
b) velocity is 3v by 4 and time is t thn
the dis is speed into t
3v/4t
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Ball is dropped from a height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = 9.8 m/s2
Final velocity of the ball = v
From second equation of motion,
time (t) taken by the ball to hit the ground can be obtained as:
S = ut + 1/2 at^2
90 = 0 + 1/2 x 9.8 x t^2
t = √18.8 = 4.29 sec
From first equation of motion,
final velocity is given as:
v = u + at = 0 + 9.8 × 4.29 = 42.04 m/s
Rebound velocity of the ball,
u(r) =(9/10 )v= (9/10)× 42.04 = 37.84 m/s
Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
v = ur + at′
0 = 37.84 + (– 9.8) t′
t' = -37.84/ -9.8 = 3.86 s
Total time taken by the ball = t + t′ = 4.29 + 3.86 = 8.15 s
As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time. The velocity with which the ball rebounds from the floor
= (9/10)× 37.84= 34.05 m/s
Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s The speed-time graph of the ball is represented in the given attachment ( please see attached file)
I hope, this will help you
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