Answer:
(i) Time taken by the ball to fall through a height of 90 m
is obtained as follows:
or
Now
From time t = 0 to t = 4.3 s,
or
In this duration speed increases
linearly with time t from 0 to during
the downward motion of the ball and this speed-time variation has been shown by
straight line OA in Fig. 3.86.
Fig. 3.86
(ii) At first collision with the
floor, speed lost by ball
.
Thus, the ball rebounds with a
speed of . For the
further upward motion, the speed at any instant t is given by
Now the speed decreases linearly
with time and becomes zero after time
Thus, the ball reaches the
highest point again after time t = 4.3 + 3.9= 8.2s from the start. Straight
line BC represents the speed-time graph for this upward motion.
(iii) At highest point, speed of
ball is zero. It again starts falling. At any instant t, its speed is given by
Again the speed of the ball
increases linearly with time t from 0 to 37.8 ms~1 (initial speed of the
previous upward motion) in the next time-interval of 3.9 s. Total time taken
from the start = 4.3 + 3.9 + 3.9 = 12.1 s. This part of motion has been shown
by straight line CD.
Here, we have assumed a negligible time
of collision between the ball and the floor.
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