Q. A body of mass $m$ thrown horizontally with velocity $v$, from the top of tower of height $h$ touches the level ground at distance of 250 m from the foot of the tower. A body of mass $2m$ thrown horizontally with velocity $\frac{v}{2}$, from the top of tower of height $4h$ will touch the level ground at a distance $x$ from the foot of tower. The value of $x$ is
A) $250\ m$
B) $500\ m$
C) $125\ m$
D) $250\sqrt{2}\ m$
Explanation
(a) $t = \sqrt{\frac{2h}{g}}$
Distance from the foot of the tower
$d = vt = v\sqrt{\frac{2h}{g}} = 250m$
When velocity $= \frac{v}{2}$
and height of tower$= 4h$
Then distance $x = \frac{v}{2}\sqrt{\frac{2(4h)}{g}}$
$x = v\sqrt{\frac{2h}{g}} = 250\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.