Question

Two balls are projected with the same speed at different angles. If the maximum height of the first ball is 2 times the maximum height of the second 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:

The time of flight is directly proportional to the sine of the angle of projection. To find the ratio, relate the maximum heights and the angles of projection.

Answer 1

The Correct Option is C

Let the speed of projection be \( u \), and the angles of projection for the two balls be \( \theta_1 \) and \( \theta_2 \). The maximum height \( H \) for a projectile is given by: \[ H = \frac{u^2 \sin^2 \theta}{2g}, \] where \( g \) is the acceleration due to gravity. For the first ball, the maximum height is: \[ H_1 = \frac{u^2 \sin^2 \theta_1}{2g}. \] For the second ball, the maximum height is: \[ H_2 = \frac{u^2 \sin^2 \theta_2}{2g}. \] We are told that \( H_1 = 2H_2 \), so: \[ \frac{u^2 \sin^2 \theta_1}{2g} = 2 \times \frac{u^2 \sin^2 \theta_2}{2g}. \] Simplifying: \[ \sin^2 \theta_1 = 2 \sin^2 \theta_2. \] This implies: \[ \sin \theta_1 = \sqrt{2} \sin \theta_2. \] Now, the time of flight \( T \) for a projectile is given by: \[ T = \frac{2u \sin \theta}{g}. \] For the first ball, the time of flight is: \[ T_1 = \frac{2u \sin \theta_1}{g}. \] For the second ball, the time of flight is: \[ T_2 = \frac{2u \sin \theta_2}{g}. \] Taking the ratio of the time of flights: \[ \frac{T_1}{T_2} = \frac{2u \sin \theta_1}{g} \div \frac{2u \sin \theta_2}{g} = \frac{\sin \theta_1}{\sin \theta_2}. \] Substituting \( \sin \theta_1 = \sqrt{2} \sin \theta_2 \), we get: \[ \frac{T_1}{T_2} = \sqrt{2}. \] Therefore, the ratio of their time of flight is \( 2 : 1 \).
Thus, the correct answer is (3) \( 2 : 1 \).