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Question 18
A ball thrown up vertically returns to the thrower after 6 s. Find

(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s

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Solution

(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey.
Hence, it has taken 3 s to attain the maximum height.
Final velocity of the ball at the maximum height, v = 0
Acceleration due to gravity, g=9.8 ms2
From the first equation of motion,
v = u + gt
0=u+(9.8×3)
u=9.8×3=29.4 ms1
Hence, the ball was thrown upwards with a velocity of 29.4 ms1

(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u=29.4 ms1
Final velocity, v = 0
Acceleration due to gravity, g=9.8 ms2
From the second equation of motion,
s=ut+12gt2
h=29.4×3+12×9.8×(3)2=44.1 m

(c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.
In this case,
Initial velocity, u = 0
Distance travelled by it during its downward journey in remaining 1 s is given by
s=0×t+12×9.8×12=4.9 m
Total height = 44.1 m
This means that the ball is 39.2 m (44.1 m - 4.9 m) above the ground after 4 seconds.

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