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 $4\, m$ 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 $h =\frac{1}{2} gt ^{2} \quad( u =0)$
$t=\frac{x}{v}$
$\therefore h =\frac{1}{2} g \left(\frac{ x ^{2}}{ v ^{2}}\right)....(1)$
$h \propto \frac{x^{2}}{v^{2}}$
$\therefore \frac{ h }{4 h }=\frac{ x ^{2}}{ v ^{2}} \times \frac{ v ^{\prime 2}}{(4 x )^{2}}$
$4v^{2} = v'^{2}$
$2v = v'$