JEE Mains · Physics · STD 11 - 3.2 motion in plane
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 :
- A \(2 \sqrt{2}: 1\)
- B \(2: 1\)
- C \(\sqrt{2}: 1\)
- D \(4: 1\)
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…
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