Question

A ball is released from the top of height 'h' metres. It takes 't' seconds to reach the ground. Where is the ball at the time t/2 s?

• A. At $\left(\frac{h}{4}\right)$ from the ground
• B. At $\left(\frac{h}{2}\right)$ from the ground
• C. At $\left(\frac{3h}{4}\right)$ from the ground
• D. Depends upon mass and volume of the ball

Answer 1

The correct option is C At $\left(\frac{3h}{4}\right)$ from the ground

Initial speed of the ball $u=0$

Distance covered by it in time $t$, $h=ut+\frac{1}{2}g{t}^2$

$\therefore$ $h=\frac{1}{2}g{t}^2$ .....(1)

Distance covered by the ball in time $t^{\prime }=\frac{t}{2}$, $H=u{t}^{\prime }+\frac{1}{2}g(t^{\prime }{)}^2$

Or $H=0+\frac{1}{2}g\frac{t^2}{4}$ .....(2)

Or $H=\frac{h}{4}$

Thus height of the ball above the ground $\Delta h=h-H$

$\therefore$ $\Delta H=h-\frac{h}{4}=\frac{3h}{4}$