Question

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 $250m$ 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. $250m$
• B. $500m$
• C. $125m$
• D. $250\sqrt{2}m$

Answer 1

The correct option is A $250m$

Time of free fall of a body from a height $h,⟹t=\sqrt{\frac{2h}{g}}$

Given, 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 , $x=\frac{v}{2}\sqrt{\frac{2\left(4h\right)}{g}}⟹x=\sqrt{\frac{2h}{g}}=250m$