Question

A body of mass $M$ thrown horizontally with velocity $v$ from the top of the tower of height $H$ touches the ground at a distance of $100 \, \text{m}$ from the foot of the tower. A body of mass $2M$ thrown at a velocity $\frac{v}{2}$ from the top of the tower of height $4H$ will touch the ground at a distance of $\dots$ $\text{m}$.

Answer 1

For the first body:
{Horizontal distance = horizontal velocity $\times$ time of flight.}
The time of flight for a height $H$ is:
$t_1 = \sqrt{\frac{2H}{g}}.$
The horizontal distance for the first body is:
$x_1 = v \times t_1 = v \times \sqrt{\frac{2H}{g}}.$
Given:
$x_1 = 100 \, \text{m}.$
For the second body:
The height is $4H$, so the time of flight is:
$t_2 = \sqrt{\frac{2(4H)}{g}} = 2\sqrt{\frac{2H}{g}}.$
The horizontal velocity is $\frac{v}{2}$. The horizontal distance for the second body is:
$x_2 = \frac{v}{2} \times t_2.$
Substitute $t_2 = 2\sqrt{\frac{2H}{g}}$:
$x_2 = \frac{v}{2} \times 2 \sqrt{\frac{2H}{g}} = v \times \sqrt{\frac{2H}{g}}.$
From the first case:
$v \times \sqrt{\frac{2H}{g}} = 100 \, \text{m}.$
Thus:
$x_2 = 100 \, \text{m}.$
{Final Result:}
$x = 100 \, \text{m}.$

Answer 2

For the first body:
{Horizontal distance = horizontal velocity $\times$ time of flight.}
The time of flight for a height $H$ is:
$t_1 = \sqrt{\frac{2H}{g}}.$
The horizontal distance for the first body is:
$x_1 = v \times t_1 = v \times \sqrt{\frac{2H}{g}}.$
Given:
$x_1 = 100 \, \text{m}.$
For the second body:
The height is $4H$, so the time of flight is:
$t_2 = \sqrt{\frac{2(4H)}{g}} = 2\sqrt{\frac{2H}{g}}.$
The horizontal velocity is $\frac{v}{2}$. The horizontal distance for the second body is:
$x_2 = \frac{v}{2} \times t_2.$
Substitute $t_2 = 2\sqrt{\frac{2H}{g}}$:
$x_2 = \frac{v}{2} \times 2 \sqrt{\frac{2H}{g}} = v \times \sqrt{\frac{2H}{g}}.$
From the first case:
$v \times \sqrt{\frac{2H}{g}} = 100 \, \text{m}.$
Thus:
$x_2 = 100 \, \text{m}.$
{Final Result:}
$x = 100 \, \text{m}.$