Question

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate the maximum height to which it rises.

Answer 1

Initial velocity (u) = 49 m/s    Final velocity (v) = 0 m/s

Gravity in toward down = + 9.8 m/s2 Gravity in toward up = - 9.8 m/s2

(i) We know, V2 - u2 = 2gs

              or,  (o)2- (49)2 = 2 X 9.8 X S

              or,  s = - (49)2/ 2 X 9.8 = 122.5 m

Maximum height = 122.5 m

(ii) We know, v = u + gt

              or   0 = 49 +(- 9.8) X t

              or   9.8 X t =  49

              or, t = 49/ 9.8 = 5 s

If, time for upward direction = time for downward direction

Then, total time taken by a ball to return back = 5 + 5 = 10 s

Answer 2

v² - u² = 2aS

Rearrange the above equation

S = (v² - u²) / (2a)
= [(0² - 49²) m²/s²] / (2 × (-9.8 m/s²))
= 122.5 m

Maximum height reached by ball is 122.5 m