Search any question & find its solution
Question:
Answered & Verified by Expert
A body of mass $5 \mathrm{~kg}$ is thrown vertically up with a kinetic energy of $490 \mathrm{~J}$. The height at which the kinetic energy of the body becomes half of the original value is
(acceleration due to gravity $=9.8 \mathrm{~ms}^{-2}$ )
Options:
(acceleration due to gravity $=9.8 \mathrm{~ms}^{-2}$ )
Solution:
2016 Upvotes
Verified Answer
The correct answer is:
$5 \mathrm{~m}$
Given, $\quad m=5 \mathrm{~kg}$
and
$\mathrm{KE}=490 \mathrm{~J}$
By the law of conservation of energy
$$
\begin{aligned}
\frac{1}{2} m u^{2} &=\frac{1}{2} m v^{2}+m g h \\
490 &=245+5 \times 9.8 \times h \\
h &=\frac{490-245}{5 \times 9.8} \\
h &=\frac{245}{49} \\
&=5 \mathrm{~m}
\end{aligned}
$$
and
$\mathrm{KE}=490 \mathrm{~J}$
By the law of conservation of energy
$$
\begin{aligned}
\frac{1}{2} m u^{2} &=\frac{1}{2} m v^{2}+m g h \\
490 &=245+5 \times 9.8 \times h \\
h &=\frac{490-245}{5 \times 9.8} \\
h &=\frac{245}{49} \\
&=5 \mathrm{~m}
\end{aligned}
$$
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.