MHT CET · Physics · Motion In Two Dimensions
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)
- A \(\frac{\mathrm{u}^2}{\mathrm{~g}}\)
- B \(\frac{\mathrm{u}^2 \cos ^2 \theta}{\mathrm{g}}\)
- C \(\frac{\mathrm{u}^2 \sin ^2 \theta}{\mathrm{g}}\)
- D \(\frac{\mathrm{u}^2 \cos ^2 \theta}{2 \mathrm{~g}}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{u}^2 \cos ^2 \theta}{\mathrm{g}}\)
Step-by-step Solution
Detailed explanation
Horizontal velocity at highest point:
\(\mathrm{v}_{\mathrm{x}}=\mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos \theta\)
\(\mathrm{a}=\frac{\mathrm{v}_{\mathrm{x}}{ }^2}{\mathrm{R}}\)
\(a=g\)
\(\therefore \quad \mathrm{R}=\frac{\mathrm{v}_{\mathrm{x}}{ }^2}{\mathrm{~g}}=\frac{(\mathrm{u} \cos \theta)^2}{\mathrm{~g}}=\frac{\mathrm{u}^2 \cos ^2 \theta}{\mathrm{g}}\)
\(\mathrm{v}_{\mathrm{x}}=\mathrm{u}_{\mathrm{x}}=\mathrm{u} \cos \theta\)
\(\mathrm{a}=\frac{\mathrm{v}_{\mathrm{x}}{ }^2}{\mathrm{R}}\)
\(a=g\)
\(\therefore \quad \mathrm{R}=\frac{\mathrm{v}_{\mathrm{x}}{ }^2}{\mathrm{~g}}=\frac{(\mathrm{u} \cos \theta)^2}{\mathrm{~g}}=\frac{\mathrm{u}^2 \cos ^2 \theta}{\mathrm{g}}\)
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