Laws of Motion — Question 11

(a) 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,

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

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,

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

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) = $\dfrac{2\text u}{\text g}$ and

u = 20 m s^-1

g = 10 m s^-2

Substituting the values in the formula, we get,

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

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