Search any question & find its solution
Question:
Answered & Verified by Expert
A stone is projected at angle ' $\theta$ ' with velocity ' $u$ '. If it executes nearly a circular motion at its maximum point for short time, the radius of the circular path will be ( $\mathrm{g}=$ acceleration due to gravity)
Options:
Solution:
1885 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{u}^2 \cos ^2 \theta}{\mathrm{g}}$
Horizontal velocity at highest point:
$\mathrm{v}_{\mathrm{x}}=\mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos \theta$
$\mathrm{a}=\frac{\mathrm{v}_{\mathrm{x}}{ }^2}{\mathrm{R}}$
$a=g$
$\therefore \quad \mathrm{R}=\frac{\mathrm{v}_{\mathrm{x}}{ }^2}{\mathrm{~g}}=\frac{(\mathrm{u} \cos \theta)^2}{\mathrm{~g}}=\frac{\mathrm{u}^2 \cos ^2 \theta}{\mathrm{g}}$
$\mathrm{v}_{\mathrm{x}}=\mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos \theta$
$\mathrm{a}=\frac{\mathrm{v}_{\mathrm{x}}{ }^2}{\mathrm{R}}$
$a=g$
$\therefore \quad \mathrm{R}=\frac{\mathrm{v}_{\mathrm{x}}{ }^2}{\mathrm{~g}}=\frac{(\mathrm{u} \cos \theta)^2}{\mathrm{~g}}=\frac{\mathrm{u}^2 \cos ^2 \theta}{\mathrm{g}}$
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.