Question

A ball is released from certain height h reaches ground in time T. Where will it be from the ground at time $\frac{3T}{4}?$

• A. $\frac{9h}{16}$
• B. $\frac{7h}{16}$
• C. $\frac{3h}{4}$
• D. $\frac{27h}{64}$

Answer 1

The correct option is B $\frac{7h}{16}$

Given that,

Height $=h$

Time $t=\frac{3T}{4}$

Since, the ball is released from rest, its initial velocity is zero. When released from height h it reaches the ground in time T.

If we apply the second equation of motion

$s=ut+\frac{1}{2}g{t}^2$

$h=\frac{g{T}^2}{2}$

$T^2=\frac{2h}{g}$

Now, we apply the same equation to calculate the height covered in time 3T/4

$s=ut+\frac{1}{2}g{t}^2$

$s=g\times \frac{{\left(\frac{3T}{4}\right)}^2}{2}$

$s=\left(\frac{9g}{32}\right)\times T^2$

$s=\left(\frac{9g}{32}\right)\times \frac{2h}{g}$

$s=\frac{9h}{16}$

Now, the distance from the ground

$s=h-\frac{9h}{16}$

$s=\frac{7h}{16}$

Hence, the distance from the ground is $\frac{7h}{16}$