(i) According to the third equation of motion under gravity:
v2−u2=2 gs
Where,
u = Initial velocity of the ball
v = Final velocity of the ball
s = Height achieved by the ball
g = Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v = 0, u = 49 m/s
During upward motion, g=−9.8 ms−2
Let h be the maximum height attained by the ball.
Hence,
0−492=2×−9.8×h
h=49×492×9.8=122.5 m
(ii) Let t be the time taken by the ball to reach the height 122.5 m, then according to the first equation of motion:
v = u + gt
We get,
0=49+t×(−9.8)
9.8t=49
t=499.8=5s
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 s + 5 s = 10 s