Question

A child is swinging a swing minimum and the maximum heights of swing from the Earth's surface are $0.75mand2m$ respectively. What will be the maximum velocity of this swing?

Answer 1

Step1: Given data:

minimum height of swing from the earth's surface( $H_1$)=$0.75m$

maximum height of swing from earth's surface ( $H_2$)=$2m$

Step 2:Formula applied

According to Energy Conservation,

$\frac{1}{2}$ ${m{v}^2}_{max}$ $=mg$$(H_2-H_1)$

Putting values of $H_1$ and $H_2$:

$\frac{1}{2}$ ${m{v}^2}_{max}$$=mg(2-0.75)$

$v_{max}$=$\sqrt{2*10*1.25}$

$v_{max}$=$\sqrt{25}$

$v_{max}$ =$5m$$s^{-1}$

Therefore, the maximum velocity will be $5m$$s^{-1}$.