A ball thrown up vertically returns to the thrower after 6 s. Find
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.
Answer
(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 m s−2
Equation of motion, v = u + gt will give,
0 = u + (−9.8 × 3)
u = 9.8 × 3 = 29.4 ms−1
Hence, the ball was thrown upwards with a velocity of 29.4 m s−1.
(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u = 29.4 m s−1
Final velocity, v = 0
Acceleration due to gravity, g = −9.8 m s−2
From the equation of motion, s= ut + 1/2 at2
h= 29.4 x 3 + 1/2 x -9.8 x (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
Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s − 3 s = 1 s.
Equation of motion, s= ut + 1/2 gt2 will give,
s= 0 x t + 1/2 x 9.8 x 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.
