Q124 / 160 · TS EAMCET 2024 (10 May Shift 1) · Physics
A body is projected from the ground at an angle of $\tan ^{-1}(\sqrt{7})$ with the horizontal. At half of the maximum height, the speed of the body is ' $n$ ' times the speed of projection. The value of ' $n$ ' is
- A2
- B$\frac{1}{3}$
- C$\frac{4}{3}$
- Correct answerD$\frac{3}{4}$
Solution
For projectile motion,
$\begin{aligned}
& \theta=\tan ^{-1}(\sqrt{7}) \Rightarrow \tan \theta=\sqrt{7} \\
& \text { At } y=\frac{H}{2}, v_y^2=u_y^2-2 g\left(\frac{H}{2}\right)
\end{aligned}$
$\begin{aligned}
& \therefore \mathrm{v}_{\mathrm{y}}^2=\mathrm{u}^2 \sin ^2 \theta-\mathrm{gH}, \mathrm{v}_{\mathrm{x}}^2=\mathrm{u}_{\mathrm{x}}^2=\mathrm{u}^2 \cos ^2 \theta \\
& \therefore \mathrm{v}=\sqrt{\mathrm{v}_{\mathrm{x}}^2+\mathrm{v}_{\mathrm{y}}^2}=\sqrt{\mathrm{u}^2\left(\sin ^2 \theta+\cos ^2 \theta\right)-\mathrm{gH}} \\
& =\sqrt{\mathrm{u}^2-\mathrm{gH}}
\end{aligned}$
Also, $v=n u \Rightarrow v^2=n^2 u^2$ $\begin{aligned} & \Rightarrow u^2-g H=n^2 u^2 \\ & \Rightarrow n^2=1-\frac{g H}{u^2}=1-\frac{\frac{u^2 \sin ^2 \theta}{2}}{u^2} \\ & =1-\frac{1}{2}\left(1-\frac{1}{1+\tan ^2 \theta}\right) \\ & =1-\frac{1}{2}\left(1-\frac{1}{1+7}\right)=\frac{9}{16} \\ & \therefore \mathrm{n}=\frac{3}{4} \end{aligned}$
Also, $v=n u \Rightarrow v^2=n^2 u^2$ $\begin{aligned} & \Rightarrow u^2-g H=n^2 u^2 \\ & \Rightarrow n^2=1-\frac{g H}{u^2}=1-\frac{\frac{u^2 \sin ^2 \theta}{2}}{u^2} \\ & =1-\frac{1}{2}\left(1-\frac{1}{1+\tan ^2 \theta}\right) \\ & =1-\frac{1}{2}\left(1-\frac{1}{1+7}\right)=\frac{9}{16} \\ & \therefore \mathrm{n}=\frac{3}{4} \end{aligned}$
Hence, the correct answer is option: 'D'.