A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19.6 m s-1. The ball reaches the ground after 5 s. Calculate: (i) the height of the tower, (ii) the velocity of ball on reaching the ground. Take g = 9.8 m s-2

(i) Initial velocity (u) = 19.6 m s^-1

Final velocity (v) = 0 (velocity on reaching the maximum height)

g = 9.8 m s^-2

Time (t) = 5 s

As we know,

v = u – gt (a = -g as the movement is against gravity)

⇒ 0 = 19.6 – 9.8 × t

⇒ 9.8 × t = 19.6

⇒ t = 19.6 / 9.8 = 2 s

Hence, the time taken to reach the maximum height is 2 s and from maximum height back to the top of the tower = 2s.

Therefore, time taken from the top of the tower to the ground = 5 – (2+2) = 1 s.

So, with the help of the formula:

h = ut + 1/2 gt²

= 19.6 × 1 + 1/2 × 9.8 × 1²

= 19.6 + 4.9 = 24.5 m

Hence, height of the tower = 24.5 m

(ii) Initial velocity (u) = 0 (on falling from maximum height)

Time taken in falling from maximum height to ground = (5 – 2) s
= 3 s

As we know, v = u + gt

Substituting the values in the formula, we get,

v = 0 + 9.8 × 3 = 29.4 ms^-1

Hence, the velocity of ball on reaching the ground = 29.4 m s ^-1