Q62 / 75 · JEE Main 2025 (8 Apr Shift 2) · Physics

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 :

  • Correct answer
    A
    $2 \sqrt{2}: 1$
  • B
    $2: 1$
  • C
    $\sqrt{2}: 1$
  • D
    $4: 1$

Solution

$\begin{aligned} & \text { Given, }\left(\mathrm{H}_{\max }\right)_1=8 \times\left(\mathrm{H}_{\max }\right)_2 \\ & \frac{\mathrm{u}^2 \sin ^2 \theta_1}{2 \mathrm{~g}}=8 \times \frac{\mathrm{u}^2 \sin ^2 \theta_2}{2 \mathrm{~g}} \\ & \Rightarrow \sin \theta_1=2 \sqrt{2} \sin \theta_2 \\ & \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\frac{2 \mathrm{u} \sin \theta_1 / \mathrm{g}}{2 \mathrm{u} \sin \theta_2 / \mathrm{g}}=\frac{\sin \theta_1}{\sin \theta_2}=2 \sqrt{2}\end{aligned}$

Hence, the correct answer is option: 'A'.

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