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

In projectile motion, the maximum height \( H \) and time of flight \( T \) are functions of initial velocity and projection angle.
The formula for maximum height is:\[H = \frac{v^2 \sin^2 \theta}{2g}\]where \( v \) denotes initial speed, \( \theta \) the projection angle, and \( g \) the acceleration due to gravity.
Let \( H_1 \) and \( H_2 \) represent the maximum heights achieved by the first and second balls, respectively. The given relationship is:\[H_1 = 8 H_2\]Substituting the maximum height formula yields:\[\frac{v^2 \sin^2 \theta_1}{2g} = 8 \times \frac{v^2 \sin^2 \theta_2}{2g}\]This simplifies to:\[\sin^2 \theta_1 = 8 \sin^2 \theta_2\]Consequently:\[\sin \theta_1 = \sqrt{8} \sin \theta_2\]The time of flight \( T \) for a projectile is calculated as:\[T = \frac{2v \sin \theta}{g}\]For the first and second balls, the times of flight are:\[T_1 = \frac{2v \sin \theta_1}{g}, \quad T_2 = \frac{2v \sin \theta_2}{g}\]The ratio of their times of flight is:\[\frac{T_1}{T_2} = \frac{\sin \theta_1}{\sin \theta_2} = \sqrt{8} = 2\sqrt{2}\]Therefore, the ratio of their times of flight is \( 2\sqrt{2} : 1 \).