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

To solve the problem, we must analyze the motion of the bodies using the principles of projectile motion. For the first case, a body of mass \( M \) is thrown horizontally from a height \( H \) with velocity \( v \) and lands 100 meters from the base of the tower.

Step 1: Determine Time of Flight for the First Case
The time of flight \( t \) for horizontal motion from height \( H \) is given by the equation: \( H = \frac{1}{2}gt^2 \). Solving for \( t \), we get \( t = \sqrt{\frac{2H}{g}} \).
Step 2: Horizontal Distance for the First Case
The horizontal distance covered is \( d = vt \). Given \( d = 100 \, \text{m} \), we have \( v\sqrt{\frac{2H}{g}} = 100 \).

Step 3: Solve for \( v \)
Rearrange to find \( v \): \( v = \frac{100}{\sqrt{\frac{2H}{g}}} \).

Step 4: Determine Flight Time for the Second Case
In the second scenario, a body of mass \( 2M \) is projected horizontally from height \( 4H \) with velocity \( \frac{v}{2} \). Using the formula \( 4H = \frac{1}{2}gt_2^2 \), solve for \( t_2 \): \( t_2 = \sqrt{\frac{8H}{g}} = 2\sqrt{\frac{2H}{g}} \).

Step 5: Calculate Horizontal Distance for the Second Case
Distance \( d_2 \) is given by \( d_2 = \left(\frac{v}{2}\right)t_2 = \left(\frac{100}{2\sqrt{\frac{2H}{g}}}\right)2\sqrt{\frac{2H}{g}} = 100 \, \text{m} \).

The solution is validated by checking the computed horizontal distance falls within the specified range, which is indeed 100 meters.

Result: The body of mass \( 2M \) touches the ground at a distance of 100 meters from the foot of the tower.

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