WBJEE · Physics · Motion In Two Dimensions
A body of mass \(m\) is thrown with velocity u from the origin of a co-ordinate axes at an angle \(\theta\) with the horizon. The magnitude of the angular momentum of the particle about the origin at time \(\mathrm{t}\) when it is at the maximum height of the trajectory is proportional to
- A \(\mathrm{u}\)
- B \(u^2\)
- C \(\mathrm{u}^3\)
- D independent of \(u\)
Answer & Solution
Correct Answer
(C) \(\mathrm{u}^3\)
Step-by-step Solution
Detailed explanation
\(\vec{L}=m u \cos \theta h_{\max }=m u \cos \theta \frac{u^2 \sin ^2 \theta}{2 g} \Rightarrow L=\frac{m u^3 \sin ^2 \theta \cos \theta}{2 g} \quad L \propto u^3\)
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