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) As we know,

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

Given,

u = 19.6 m s^-1

v = 0 (velocity on reaching the maximum height)

g = 9.8 m s^-2

t = 5 s

Substituting the values in the formula, we get the time taken to reach the maximum height.

$0 = 19.6 - 9.8 \times t \\[0.5em] 19.6 = 9.8 \times t \\[0.5em] \Rightarrow t = \dfrac{19.6}{9.8} \\[0.5em] \Rightarrow t = 2 s \\[0.5em]$

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 + $\dfrac{1}{2}$ gt²

we get,

$h = (19.6 \times 1) + (\dfrac{1}{2} \times 9.8\times 1^2) \\[0.5em] h = 19.6 + 4.9 \\[0.5em] h = 24.5 m \\[0.5em]$

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 = Total Time - Time taken to reach maximum height
= (5 - 2) s
= 3 s

As we know, v = u + gt

Substituting the values in the formula, we get,

$v = 0 + (9.8 \times 3) \\[0.5em] \Rightarrow v = 29.4 m s ^{-1} \\[0.5em]$

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