Q123 / 160 · AP EAMCET 2023 (15 May Shift 1) · Physics
A stone thrown with velocity ' $u$ ' at angles ' $\theta^{\prime}$ and $\left(90^{\circ}-\theta\right)$ with the horizontal reaches to maximum heights $\mathrm{H}_1$ and $\mathrm{H}_2$ respectively. Its horizontal range is
- Correct answerA$4 \sqrt{\mathrm{H}_1 \mathrm{H}_2}$
- B$2 \mathrm{H}_1 \mathrm{H}_2$
- C$2 \sqrt{\mathrm{H}_1 \mathrm{H}_2}$
- D$\sqrt[4]{\frac{\mathrm{H}_1}{\mathrm{H}_2}}$
Solution
Maximum height, $H=\frac{u^2 \sin ^2 \theta}{2 g}$ and horizontal range,
$$
R=\frac{u^2 \sin 2 \theta}{g}
$$
$\begin{aligned} & \mathrm{H}_1=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \text { and } \mathrm{H}_2=\frac{\mathrm{u}^2 \sin ^2\left(90^{\circ}-\theta\right)}{2 \mathrm{~g}}=\frac{\mathrm{u}^2 \cos ^2 \theta}{2 \mathrm{~g}} \\ & \therefore \mathrm{H}_1 \mathrm{H}_2=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \times \frac{\mathrm{u}^2 \cos ^2 \theta}{2 \mathrm{~g}}=\frac{\left(\mathrm{u}^2 \sin 2 \theta\right)^2}{16 \mathrm{~g}^2}=\frac{\mathrm{R}^2}{16} \\ & \therefore \mathrm{R}=4 \sqrt{\mathrm{H}_1 \mathrm{H}_2}\end{aligned}$
Hence, the correct answer is option: 'A'.