Question

A body of mass $m$ is projected horizontally with a velocity $v$ from the top of a tower of height $h$ and it reaches the ground at a distance $x$ from the foot of the tower. If a second body of mass $2m$ is projected horizontally from the top of a tower of height $2h$, it reaches the ground at a distance $2x$ from the foot of the tower. The horizontal velocity of the second body is-

• A. $v$
• B. $2v$
• C. $\sqrt{2}v$
• D. $\frac{v}{2}$

Answer 1

Answer: (C)

$\sqrt{2}v$

$\text{Time taken to cover height h}=\sqrt{\frac{2h}{g}}$
$\text{Distance covered in same time}x=vt=v\sqrt{\frac{2h}{g}}$
$\text{Time taken to cover height 2h}=\sqrt{\frac{4h}{g}}$
$\text{Distance covered in same time}2x=vt=v\sqrt{\frac{4h}{g}}$
$\Rightarrow 2V\sqrt{\frac{2h}{g}}=v\sqrt{\frac{4h}{g}}$
$\Rightarrow V=\sqrt{2}v$