Question

A ball thrown up, vertically returns to the thrower after 6 seconds find:

(a) the velocity with which it was thrown up

(b) the maximum height it reaches

(c) its position after 4 seconds

Answer 1

(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=−9.8ms−2
From the first equation of motion,
v = u + gt
0=u+(−9.8×3)
u=9.8×3=29.4ms−1
Hence, the ball was thrown upwards with a velocity of 29.4ms−1

(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u=29.4ms−1
Final velocity, v = 0
Acceleration due to gravity, g=−9.8ms−2
From the second equation of motion,
s=ut+12gt2
h=29.4×3+12×−9.8×(3)2=44.1m

(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=4.9m
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.