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

Correct Answer: 100

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:}
\[x = 100 \, \text{m}.\]