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 projection speed be \( u \) and the projection angles for the two balls be \( \theta_1 \) and \( \theta_2 \). The maximum height \( H \) of a projectile is \( H = \frac{u^2 \sin^2 \theta}{2g} \), where \( g \) is the acceleration due to gravity.For the first ball, \( H_1 = \frac{u^2 \sin^2 \theta_1}{2g} \). For the second ball, \( H_2 = \frac{u^2 \sin^2 \theta_2}{2g} \).Given \( H_1 = 2H_2 \), we have \( \frac{u^2 \sin^2 \theta_1}{2g} = 2 \times \frac{u^2 \sin^2 \theta_2}{2g} \), which simplifies to \( \sin^2 \theta_1 = 2 \sin^2 \theta_2 \), or \( \sin \theta_1 = \sqrt{2} \sin \theta_2 \).The time of flight \( T \) for a projectile is \( T = \frac{2u \sin \theta}{g} \).For the first ball, \( T_1 = \frac{2u \sin \theta_1}{g} \). For the second ball, \( T_2 = \frac{2u \sin \theta_2}{g} \).The ratio of the time of flights is \( \frac{T_1}{T_2} = \frac{2u \sin \theta_1}{g} \div \frac{2u \sin \theta_2}{g} = \frac{\sin \theta_1}{\sin \theta_2} \).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 \( \sqrt{2} : 1 \). This corresponds to a ratio of \( 2 : 1 \) when considering the common convention of expressing ratios with integers. The correct answer is (3) \( 2 : 1 \).