KCET · Physics · Motion In Two Dimensions
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 \(3:1\)
- B \(1:2\)
- C \(4:1\)
- D \(1:3\)
Answer & Solution
Correct Answer
(B) \(1:2\)
Step-by-step Solution
Detailed explanation
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: \(1:2\)
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