Q124 / 160 · TS EAMCET 2024 (11 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
- A$2$
- B$\frac{1}{2}$
- C$\frac{4}{3}$
- Correct answerD$\frac{3}{4}$
Solution
For projectile motion,
$\begin{aligned}
& \theta=\tan ^{-1}(\sqrt{7}) \Rightarrow \tan \theta=\sqrt{7} \\
& \text { At } y=\frac{H}{2}, v_y^2=u_y^2-2 g\left(\frac{H}{2}\right) \\
& \therefore \quad v_y^2=u^2 \sin ^2 \theta-g H, v_x^2=u_x^2=u^2 \cos ^2 \theta \\
& \therefore \quad v=\sqrt{v_x^2+v_y^2}=\sqrt{u^2\left(\sin ^2 \theta+\cos ^2 \theta\right)-g H} \\
& =\sqrt{u^2-\mathrm{gH}}
\end{aligned}$
Also, $v=\mathrm{nu} \Rightarrow \mathrm{v}^2=\mathrm{n}^2 \mathrm{u}^2$ $\begin{aligned} & \Rightarrow \mathrm{u}^2-\mathrm{gH}=\mathrm{n}^2 \mathrm{u}^2 \\ & \Rightarrow \mathrm{n}^2=1-\frac{\mathrm{gH}}{\mathrm{u}^2}=1-\frac{\frac{\mathrm{u}^2 \sin ^2 \theta}{2}}{\mathrm{u}^2}=1-\frac{1}{2}\left(1-\frac{1}{1+\tan ^2 \theta}\right) \\ & =1-\frac{1}{2}\left(1-\frac{1}{1+7}\right)=\frac{9}{16} \therefore \mathrm{n}=\frac{3}{4} \end{aligned}$
Also, $v=\mathrm{nu} \Rightarrow \mathrm{v}^2=\mathrm{n}^2 \mathrm{u}^2$ $\begin{aligned} & \Rightarrow \mathrm{u}^2-\mathrm{gH}=\mathrm{n}^2 \mathrm{u}^2 \\ & \Rightarrow \mathrm{n}^2=1-\frac{\mathrm{gH}}{\mathrm{u}^2}=1-\frac{\frac{\mathrm{u}^2 \sin ^2 \theta}{2}}{\mathrm{u}^2}=1-\frac{1}{2}\left(1-\frac{1}{1+\tan ^2 \theta}\right) \\ & =1-\frac{1}{2}\left(1-\frac{1}{1+7}\right)=\frac{9}{16} \therefore \mathrm{n}=\frac{3}{4} \end{aligned}$
Hence, the correct answer is option: 'D'.