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 $4m$ is projected horizontally from the top of a tower of height $4h$, it reaches the ground at a distance $4x$ from the foot of the tower. The horizontal velocity of the second body is:

• A. $6V$
• B. $\sqrt{2V}$
• C. $2V$
• D. $5V$

Answer 1

Answer: (C)

For the first body $h=\frac{1}{2}g{t}^2(u=0)$
$t=\frac{x}{v}$
$\therefore h=\frac{1}{2}g\left(\frac{x^2}{v^2}\right)....(1)$
$h\propto \frac{x^2}{v^2}$
$\therefore \frac{h}{4h}=\frac{x^2}{v^2}\times \frac{v^{\prime 2}}{(4x{)}^2}$
$4v^2=v^{\prime 2}$
$2v=v^{\prime }$