JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2025
- Easy
- MCQ
Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. \(T_1\) and \(T_2\) are the total flying times of first and second ball, respectively, then the ratio of \(T_1\) and \(T_2\) is :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(2 \sqrt{2}: 1\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Given, }\left(\mathrm{H}_{\max }\right)_1=8 \times\left(\mathrm{H}_{\max }\right)_2 \\ & \frac{\mathrm{u}^2 \sin ^2 \theta_1}{2 \mathrm{~g}}=8 \times \frac{\mathrm{u}^2 \sin ^2 \theta_2}{2 \mathrm{~g}} \\ & \Rightarrow \sin \theta_1=2 \sqrt{2} \sin \theta_2 \\ & \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\frac{2 \mathrm{u} \sin \theta_1 / \mathrm{g}}{2 \mathrm{u} \sin \theta_2 / \mathrm{g}}=\frac{\sin \theta_1}{\sin \theta_2}=2 \sqrt{2}\end{aligned}\)
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