A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Answer:
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
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:
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, ur =
Time (t) taken by the
ball to reach maximum height is obtained with the help of first equation of motion as:
v = ur + at′
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
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 figure as:
The speed-time graph of the ball is represented in the given figure as:
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