A ball is thrown vertically upwards. It goes to a height 20 m and then returns to the ground. Taking acceleration due to gravity g to be 10 m s^-2

find —

(a) the initial velocity of the ball

(b) the final velocity of the ball on reaching the ground and

(c) the total time of journey of the ball.

As we know, from the equation of motion,

v² = u² - 2gs (a = - g as movement is against gravity )

Given,

s = 20 m

g = 10 m s^-2

v = 0

Substituting the values in the formula, we get,

$0 = u^2 - 2 \times 10 \times 20 \\[0.5em] u^2 = 400 \\[0.5em] \Rightarrow u = 20 \text{m s}^{-1} \\[0.5em]$

Hence, initial velocity of the ball = 20 m s^-1

(b) As we know, from the equation of motion,

v² = u² + 2gs

When the ball starts falling after reaching the maximum height, it's velocity (u) = 0

s = 20 m

g = 10 m s^-2

Substituting the values in the formula, we get,

$v^2 = 0^2 + (2 \times 10 \times 20) \\[0.5em] v^2 = 400 \\[0.5em] \Rightarrow v = 20 \text{m s}^{-1} \\[0.5em]$

Hence, final velocity of the ball on reaching the ground = 20 m s^-1

(c) As we know,

total time of of journey of ball (t) = $\dfrac{2u}{g}$ and

u = 20 m s^-1

g = 10 m s^-2

Substituting the values in the formula, we get,

$t = \dfrac{2 \times 20}{10} \\[0.5em] \Rightarrow t = 4 s \\[0.5em]$

Hence, total time for which the ball stays in air = 4 s