Search any question & find its solution
Question:
Answered & Verified by Expert
A bullet is fired at time $t=0$ with velocity $20 \mathrm{~m} / \mathrm{s}$ and at an initial angle of $30^{\circ}$ with the horizontal. The an angle between the displacement vector and the horizontal after time $0.1 \mathrm{~s}$ is
(Assume $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
Options:
(Assume $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
Solution:
2139 Upvotes
Verified Answer
The correct answer is:
$\frac{19}{20 \sqrt{3}}$
$x=20 \cos 30^{\circ} \times 0.1=2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3} \mathrm{~m}$
$$
\begin{aligned}
& \text { and, } y=20 \sin 30^{\circ} \times 0.1-\frac{1}{2} \mathrm{~g}(0.1)^2 \\
& =1-0.05=0.95 \mathrm{~m}=\frac{19}{20} \mathrm{~m}
\end{aligned}
$$
So, $\tan \theta=\frac{y}{x}=\frac{19}{20 \sqrt{3}}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.