Q130 / 160 · TS EAMCET 2018 (07 May Shift 1) · Physics
A $1.5 \mathrm{~kg}$ ball is shot upward at an angle of $34^{\circ}$ to the horizontal with an initial speed of $20 \mathrm{~m} / \mathrm{s}$, then maximum height of the ball reaches is (use $\cos 34^{\circ}=0.83$ or $\sin 34^{\circ}=0.56$ )
- Correct answerA$6.3 \mathrm{~m}$
- B$9.4 \mathrm{~m}$
- C$13.8 \mathrm{~m}$
- D$11.2 \mathrm{~m}$
Solution
Given, inífial speed, $\mathrm{u}=20 \mathrm{~m} / \mathrm{s}$
angle of projection, $\theta=34^{\circ}$
Maximum height, $H=\frac{\mu^2 \sin ^2 \theta}{2 g}$
$$
\begin{aligned}
& r=\frac{(20)^2 \times \sin ^2 34}{2 \times 9.8}=\frac{400 \times\langle 0.56)^2}{2 \times 9.8} \\
& H=\frac{125.44}{19.6}=6.3 \mathrm{~m}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.