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

Answer: (A)

$\frac{\sqrt3}{\sqrt2}$

Answer 2

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:

$\frac{v_A}{v_B} = \sqrt{\frac{3}{2}}.$