Question

Two particles are projected from a tower of height 400 m & angles 45° & 60° horizontally. If they have the same time of flight, find the ratio of their velocities.

• A. \(\frac{\sqrt3}{\sqrt2}\)
• B. \(\frac{\sqrt5}{\sqrt2}\)
• C. \(\frac{\sqrt3}{\sqrt4}\)
• D. 1

Answer 1

The Correct Option is A

To analyze the problem, we need to consider the equations of motion for both projectiles.

The time of flight T for a projectile launched from a height h is given by:

\( T = \frac{2v \sin(\theta)}{g} + \sqrt{\frac{2h}{g}}, \)

where v is the initial speed of projection, \(\theta\) is the angle of projection with the horizontal, g is the acceleration due to gravity, and h is the height of the launch point above the ground.

For both projectiles, the total height of the tower is \( h = 400 \, \text{m} \).

For Projectile A (\(\theta = 45^\circ\)):

\( T_A = \frac{2v_A \sin(45^\circ)}{g} + \sqrt{\frac{2 \times 400}{g}}. \)

Since \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\), this becomes:

\( T_A = \frac{2v_A \cdot \frac{\sqrt{2}}{2}}{g} + \sqrt{\frac{800}{g}} = \frac{v_A \sqrt{2}}{g} + \sqrt{\frac{800}{g}}. \)

For Projectile B (\(\theta = 60^\circ\)):

\( T_B = \frac{2v_B \sin(60^\circ)}{g} + \sqrt{\frac{2 \times 400}{g}}. \)

Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), this becomes:

\( T_B = \frac{2v_B \cdot \frac{\sqrt{3}}{2}}{g} + \sqrt{\frac{800}{g}} = \frac{v_B \sqrt{3}}{g} + \sqrt{\frac{800}{g}}. \)

Equating the Times of Flight:

It is given that \( T_A = T_B \), so:

\( \frac{v_A \sqrt{2}}{g} + \sqrt{\frac{800}{g}} = \frac{v_B \sqrt{3}}{g} + \sqrt{\frac{800}{g}}. \)

Cancelling \(\sqrt{\frac{800}{g}}\) from both sides:

\( \frac{v_A \sqrt{2}}{g} = \frac{v_B \sqrt{3}}{g}. \)

Simplify:

\( v_A \sqrt{2} = v_B \sqrt{3}. \)

Finding the Ratio of Speeds: Rearranging for \(\frac{v_A}{v_B}\):

\( \frac{v_A}{v_B} = \frac{\sqrt{3}}{\sqrt{2}}. \)

This can be written as:

\( \frac{v_A}{v_B} = \sqrt{\frac{3}{2}}. \)