Question

A 2 kg ball is thrown vertically upward and another 3 kg ball is projected with a certain angle (\( \theta \neq 90^\circ \)). Both will have the same time of flight. The ratio of their maximum heights is:

• A. \( 2:3 \)
• B. \( 3:2 \)
• C. \( \sqrt{3} : 2 \)
• D. \( 1:1 \)

Hint:

For objects having the same time of flight, their maximum heights depend only on their vertical components, making the ratio 1:1.

Answer 1

The Correct Option is D

Step 1: Determine the Time of Flight Formula For vertical motion, time of flight is given by: \[ T = \frac{2 u}{g} \] For projectile motion at an angle \( \theta \): \[ T = \frac{2 u \sin \theta}{g} \] Since both objects have the same time of flight: \[ \frac{2 u}{g} = \frac{2 u \sin \theta}{g} \] Cancel \( g \) and \( 2 \), giving: \[ u = u \sin \theta \] Step 2: Find the Maximum Height Ratio For vertical motion: \[ H_1 = \frac{u^2}{2g} \] For projectile motion: \[ H_2 = \frac{(u \sin \theta)^2}{2g} \] Since \( u = u \sin \theta \), we get: \[ H_1 = H_2 \] Thus, the ratio is: \[ 1:1 \]