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 resolve the issue, projectile motion principles are applied to analyze the bodies' movement. In the initial scenario, a body of mass \( M \) is launched horizontally from an altitude of \( H \) with speed \( v \), landing 100 meters from the tower's base.

Step 1: Calculate Time of Flight (Scenario 1)
The time of flight \( t \) for horizontal projection from height \( H \) is determined by \( H = \frac{1}{2}gt^2 \). Solving for \( t \) yields \( t = \sqrt{\frac{2H}{g}} \).
Step 2: Determine Horizontal Range (Scenario 1)
The horizontal distance covered is \( d = vt \). Given \( d = 100 \, \text{m} \), we have the equation \( v\sqrt{\frac{2H}{g}} = 100 \).

Step 3: Isolate Initial Velocity \( v \)
Rearranging the equation provides \( v = \frac{100}{\sqrt{\frac{2H}{g}}} \).

Step 4: Calculate Time of Flight (Scenario 2)
For the second case, a body of mass \( 2M \) is projected horizontally from an altitude of \( 4H \) with speed \( \frac{v}{2} \). Using the equation \( 4H = \frac{1}{2}gt_2^2 \), we solve for \( t_2 \): \( t_2 = \sqrt{\frac{8H}{g}} = 2\sqrt{\frac{2H}{g}} \).

Step 5: Compute Horizontal Range (Scenario 2)
The horizontal distance \( d_2 \) is calculated as \( d_2 = \left(\frac{v}{2}\right)t_2 = \left(\frac{100}{2\sqrt{\frac{2H}{g}}}\right)2\sqrt{\frac{2H}{g}} = 100 \, \text{m} \).

The computed horizontal distance of 100 meters confirms the solution within the specified parameters.

Conclusion: The body of mass \( 2M \) impacts the ground at a distance of 100 meters from the tower's base.