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: $$1:1$$