Question

A player throws a ball upwards with an initial speed of 29.4 ms^-1. The height to which the ball rises and the time taken to reach the player's hands are

• A. 22.05 m, 38 s
• B. 44.1 m, 6s`
• C. 29.4 m, 6s
• D. 54.5 m, 9s

Answer 1

Answer: (B)

44.1 m, 6s`

Answer 2

From u = $\sqrt { 2 g h }$ h = $\frac { u ^ { 2 } } { 2 g }$ = 44.1 m

h' = ut + 1/2 gt² = 29.4 t – 1/2 (9.8)t² = 0

∴ t = 6 s

Using g = u/t, where t is time taken to reach the highest point, we get

t = u/g = 29.4/9.8 = 3 s

By symmetry, the time taken by the ball to reach from the highest point to the hands of the player is 3s.

∴ total time = 3 + 3 = 6s.