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. $\sqrt{2v}$
• D. $v/2$

Answer 1

Answer: (C)

For the first body, $h=\frac{1}{2} gt^{2}\,\,\,\,\,\,(i)$
and $x=vt \,\,\,\,\,\,\,(ii)$
From equation (i) and (ii)
$h=\frac{1}{2}g. (\frac{x^{2}}{v^{2}}) \,\,\,\,\,\,\,\,(iii)$
Dividing equation (iii) by equation (iv) we get
$\frac{1}{2}= \frac{x^{2}}{v^{2}} \times \frac{v'^{2}}{4x^{2}}$ or $v' =\sqrt{2v}$