Search any question & find its solution
Question:
Answered & Verified by Expert
A body is projected from the ground with a velocity $\mathbf{v}=(3 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}) \mathrm{ms}^{-1} .$ The maximum height attained and the range of the body respectively are (given $g=10 \mathrm{ms}^{-2}$ )
Options:
Solution:
2363 Upvotes
Verified Answer
The correct answer is:
$5 \mathrm{m}$ and $6 \mathrm{m}$
(a) Given, $v=(3 i+10 j) m / s$
$\Rightarrow \quad v_{x}=3$ and $v_{y}=10$
$\therefore$ Maximum height attained, $H=\frac{v_{y}^{2}}{2 g}$
$$
\begin{aligned}
&=\frac{10 \times 10}{2 \times 10}=5 \mathrm{m} \\
\text { Range }=v_{x} \times T=V_{x} \times \frac{2 v_{y}}{g} &=\frac{3 \times 2 \times 10}{10}=6 \mathrm{m}
\end{aligned}
$$
$\Rightarrow \quad v_{x}=3$ and $v_{y}=10$
$\therefore$ Maximum height attained, $H=\frac{v_{y}^{2}}{2 g}$
$$
\begin{aligned}
&=\frac{10 \times 10}{2 \times 10}=5 \mathrm{m} \\
\text { Range }=v_{x} \times T=V_{x} \times \frac{2 v_{y}}{g} &=\frac{3 \times 2 \times 10}{10}=6 \mathrm{m}
\end{aligned}
$$
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.