Q. A body of mass $m$ thrown horizontally with velocity $v$from the top of tower of height$\ h$, touches the level ground a distance of 250 m from the foot of the tower. A body of mass, $2\ m$ thrown horizontally with velocity $\frac{v}{2},$ from the top of tower of height $4\ h$ 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) Time, $t = \sqrt{\frac{2h}{g}}$
Distance from the foot of the tower
$\ \ \ \ \ \ \ \ \ \ \ d = vt = v\ \sqrt{\frac{2h}{g}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 250\ m$
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.