Q215 / 240 · KCET 2026 · Physics
Two bodies are projected with the same velocity. If one is projected at an angle of $30^\circ$ and the other at $45^\circ$ to the horizontal, then the ratio of maximum heights attained is
- A$3:1$
- Correct answerB$1:2$
- C$4:1$
- D$1:3$
Solution
The maximum height attained by a projectile is given by $H = \dfrac{u^2 \sin^2 \theta}{2g}$.
For the first body, $\theta_1 = 30^\circ$:
$H_1 = \dfrac{u^2 \sin^2 30^\circ}{2g} = \dfrac{u^2}{2g} \left(\dfrac{1}{2}\right)^2 = \dfrac{u^2}{8g}$
For the second body, $\theta_2 = 45^\circ$:
$H_2 = \dfrac{u^2 \sin^2 45^\circ}{2g} = \dfrac{u^2}{2g} \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{u^2}{4g}$
The ratio of maximum heights is:
$\dfrac{H_1}{H_2} = \dfrac{\dfrac{u^2}{8g}}{\dfrac{u^2}{4g}} = \dfrac{4}{8} = \dfrac{1}{2}$
Thus, the ratio is $1:2$.
Answer: $1:2$
For the first body, $\theta_1 = 30^\circ$:
$H_1 = \dfrac{u^2 \sin^2 30^\circ}{2g} = \dfrac{u^2}{2g} \left(\dfrac{1}{2}\right)^2 = \dfrac{u^2}{8g}$
For the second body, $\theta_2 = 45^\circ$:
$H_2 = \dfrac{u^2 \sin^2 45^\circ}{2g} = \dfrac{u^2}{2g} \left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{u^2}{4g}$
The ratio of maximum heights is:
$\dfrac{H_1}{H_2} = \dfrac{\dfrac{u^2}{8g}}{\dfrac{u^2}{4g}} = \dfrac{4}{8} = \dfrac{1}{2}$
Thus, the ratio is $1:2$.
Answer: $1:2$
Hence, the correct answer is option: 'B'.