Question

A body of mass $m$ thrown horizontally with velocity $v$, from the top of a tower of height $h$ touches the level ground at a distance of $250\text{m}$ from the foot of the tower. A body of mass $2m$ thrown horizontally with velocity $\frac{v}{2}$ from the top of a tower of height $4h$ will touch the level ground at a distance $x$ from the foot of tower. The value of $x$ is

• A. $250\text{m}$
• B. $500\text{m}$
• C. $125\text{m}$
• D. $250\sqrt{2}\text{m}$

Answer 1

The correct option is A $250\text{m}$

Time taken by the body to reach the ground $t=\sqrt{\frac{2h}{g}}$
Distance at which the body of mass $m$ will strike the ground from the foot of the tower.
$x=vt=v\sqrt{\frac{2h}{g}}=250\text{m}$
Distance at which the body of mass $4m$ will strike the ground from the foot of the tower
$x^{\prime }=\frac{v}{2}\sqrt{\frac{2\left(4h\right)}{g}}=x=250\text{m}$

Alternative Solution:

Horizontal distance covered by a body falling from height $h$ with initial horizontal velocity $u$
$X=u\sqrt{\frac{2h}{g}}\therefore \frac{X_2}{X_1}=\frac{u_2}{u_1}\sqrt{\frac{h_2}{h_1}}=\frac{\frac{v}{2}}{v}\sqrt{\frac{4h}{h}}=1\Rightarrow \frac{X_2}{250}=1\Rightarrow X_2=X_1=250\text{m}$