Q126 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Physics
A projectile object is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal with velocity of $140 \mathrm{~m} / \mathrm{s}$. Then, the time after which its velocity makes an angle $30^{\circ}$ with the horizontal is (Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
- Correct answerA$\frac{14}{\sqrt{3}} S$
- B$7 \sqrt{3} \mathrm{~s}$
- C$14 \sqrt{3} \mathrm{~s}$
- D$\frac{7}{\sqrt{3}} \mathrm{~s}$
Solution
Velocity of the projectile, $v=140 \mathrm{~m} / \mathrm{s}$
The angle of projection, $\theta=60^{\circ}$. The horizontal component of the velocity is always constant throughout the motion.
Let the velocity is $u$ when the angle is $30^{\circ}$ with horizontal.
$$
\begin{array}{ll}
\therefore & v \cos 60^{\circ}=u \cos 30^{\circ} \\
\Rightarrow & \frac{v}{2}=\frac{\sqrt{3} u}{2} \Rightarrow u=\frac{v}{\sqrt{3}}=\frac{140}{\sqrt{3}} \mathrm{~m} / \mathrm{s}
\end{array}
$$
For vertical motion, $u_y-v_y=g t$
$$
\begin{array}{ll}
\Rightarrow & v \sin 60^{\circ}-u \sin 30^{\circ}=g t \\
\Rightarrow & \frac{\sqrt{3} v}{2}-\frac{140}{\sqrt{3}} \times \frac{1}{2}=g t \\
\Rightarrow & g t=\frac{140 \sqrt{3}}{2}-\frac{140}{2 \sqrt{3}}=\frac{140}{2}\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) \\
\Rightarrow & t=\frac{70(3-1)}{\sqrt{3} \times 10}=\frac{14}{\sqrt{3}} \mathrm{~s} \quad\left(\because g=10 \mathrm{~m} / \mathrm{s}^2\right)
\end{array}
$$
Hence, the correct answer is option: 'A'.