Time taken to reach the maximum height

t = 6/2

t = 3s

v = 0

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(a) u = ?

from the equation, v = u – gt

0 = u - 9.8 x 3

-u = -29.4

u = 29.4 m/s

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(b) from the equation. v² = u² – 2gh

0 =  (29.4)² – 2 x 9.8 x h

19.6h = 29.4 x 29.4

h = (29.4 x 29.4) / 19.6

h = 44.1 m

Thus the maximum height reached by the ball, 44.1 m

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(c) t = 4s (given)

but the ball reaches the maximum height after 3s. Therefore the distance covered after falling from the maximum height after 1s so t = 1s.

h = ut + 0.5 gt²‍

h = ( 0 x 1) + (0.5 x 9.8 x 1 x 1)

h = 0 + 4.9

h = 4.9 ~ 5m

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∴ Height above the ground, H = 44.1 - 4.9

H = 39.2 ~ 39m

Hence after 4 s, the ball is at a height of 39 m above the ground.