Question

A body is dropped from a certain height 'h' meters. Assuming that the gravitational field is nullified, after the body has travelled $\frac{h}{2}$ meters such that g $=0,$discuss the motion of the body. Find an expression for the time taken by the body to reach the ground.

• A. $\frac{3}{2}\sqrt{\frac{h}{g}}$
• B. $\sqrt{\frac{g}{h}}$
• C. $\sqrt{gh}$
• D. $\frac{1}{2}\sqrt{gh}$

Answer 1

The correct option is A $\frac{3}{2}\sqrt{\frac{h}{g}}$

Till $\frac{h}{2}$ meter,

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

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2\times \frac{h}{2}}{g}}=\sqrt{\frac{h}{g}}$

so to cover $\frac{h}{2}$ meter with a=g,the time taken $t_1=\sqrt{\frac{h}{g}}$

After that a=0

$v(att=\sqrt{\frac{h}{g}})=g{t}_1=g\sqrt{\frac{h}{g}}=\sqrt{gh}$

with this contant speed,particle will travel next $\frac{h}{2}$ meter.

$t_2=\frac{\frac{h}{2}}{\sqrt{gh}}=\frac{1}{2}\sqrt{\frac{h}{g}}t=t_1+t_2=\sqrt{\frac{h}{g}}+\frac{1}{2}\sqrt{\frac{h}{g}}=\frac{3}{2}\sqrt{\frac{h}{g}}$