Question

Two balls are projected with the same speed at different angles. If the maximum height of the 1st ball is 8 times the maximum height of the 2nd ball, find the ratio of their time of flight.

• A. 1 : \( 2\sqrt{2} \)
• B. \( 2\sqrt{2} \) : 1
• C. 2 : 1
• D. 4 : 1

Hint:

For projectile motion, the maximum height is proportional to the square of the sine of the angle of projection, and the time of flight is proportional to the sine of the angle.

Answer 1

The Correct Option is B

For projectile motion, the maximum height \( H \) and time of flight \( T \) are related to the initial velocity and angle of projection.
The maximum height is given by: \[ H = \frac{v^2 \sin^2 \theta}{2g} \] where \( v \) is the initial speed, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity. Let \( H_1 \) and \( H_2 \) be the maximum heights of the first and second balls, respectively.
The given condition is: \[ H_1 = 8 H_2 \] Using the formula for maximum height: \[ \frac{v^2 \sin^2 \theta_1}{2g} = 8 \times \frac{v^2 \sin^2 \theta_2}{2g} \] Simplifying: \[ \sin^2 \theta_1 = 8 \sin^2 \theta_2 \] Thus: \[ \sin \theta_1 = \sqrt{8} \sin \theta_2 \] Next, the time of flight \( T \) for a projectile is given by: \[ T = \frac{2v \sin \theta}{g} \] For the first and second balls, the times of flight are: \[ T_1 = \frac{2v \sin \theta_1}{g}, \quad T_2 = \frac{2v \sin \theta_2}{g} \] Taking the ratio of their times of flight: \[ \frac{T_1}{T_2} = \frac{\sin \theta_1}{\sin \theta_2} = \sqrt{8} = 2\sqrt{2} \] Thus, the ratio of their time of flight is \( 2\sqrt{2} : 1 \).