Search any question & find its solution
Question:
Answered & Verified by Expert
A stone is projected vertically upwards with speed ' $v$ '. Another stone of same mass is projected at an angle of $60^{\circ}$ with the vertical with the same speed ' $v$ '. The ratio of their potential energies at the highest points of their journey is $\left[\sin 30^{\circ}=\cos 60^{\circ}=0.5\right.$, $\left.\cos 30^{\circ}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]$
Options:
Solution:
2526 Upvotes
Verified Answer
The correct answer is:
$4:1$
P.E. of the stone projected vertically is,
P.E. = mgh
But $\mathrm{h}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}$
$\begin{aligned}
\therefore \quad \text { P. } \mathrm{E}_1 & =\mathrm{mg}\left(\frac{\mathrm{v}^2}{2 \mathrm{~g}}\right) \\
& =\frac{\mathrm{mv}^2}{2}... (i)
\end{aligned}$
For the second stone thrown at an angle $\theta$ to the horizontal,
$\begin{array}{ll}
& \mathrm{h}=\frac{\mathrm{v}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{\mathrm{v}^2 \sin ^2 30^{\circ}}{2 \mathrm{~g}}=\frac{\mathrm{v}^2}{8 \mathrm{~g}} \\
\therefore \quad & \text { P.E. } 2=\mathrm{mg}\left(\frac{\mathrm{v}^2}{8 \mathrm{~g}}\right)=\frac{\mathrm{mv}^2}{8}... (ii)
\end{array}$
Dividing equation (i) by equation (ii) -
$-\frac{P \cdot E_1}{P \cdot E_2}=\frac{\left(\frac{m v^2}{2}\right)}{\left(\frac{m v^2}{8}\right)}=4: 1$
P.E. = mgh
But $\mathrm{h}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}$
$\begin{aligned}
\therefore \quad \text { P. } \mathrm{E}_1 & =\mathrm{mg}\left(\frac{\mathrm{v}^2}{2 \mathrm{~g}}\right) \\
& =\frac{\mathrm{mv}^2}{2}... (i)
\end{aligned}$
For the second stone thrown at an angle $\theta$ to the horizontal,
$\begin{array}{ll}
& \mathrm{h}=\frac{\mathrm{v}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{\mathrm{v}^2 \sin ^2 30^{\circ}}{2 \mathrm{~g}}=\frac{\mathrm{v}^2}{8 \mathrm{~g}} \\
\therefore \quad & \text { P.E. } 2=\mathrm{mg}\left(\frac{\mathrm{v}^2}{8 \mathrm{~g}}\right)=\frac{\mathrm{mv}^2}{8}... (ii)
\end{array}$
Dividing equation (i) by equation (ii) -
$-\frac{P \cdot E_1}{P \cdot E_2}=\frac{\left(\frac{m v^2}{2}\right)}{\left(\frac{m v^2}{8}\right)}=4: 1$
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.