JEE Mains · Physics · STD 11 - 3.2 motion in plane
A ball of mass \(160\, g\) is thrown up at an angle of \(60^o\) to the horizontal at a speed of \(10\, m\,s^{-1}\) . The angular momentum of the ball at the highest point of the trajectcry with respect to the point from which the ball is thrown is nearly ........ \(kg\, m^2/s\) \((g\, = 10\, m\,s^{-2})\)
- A \(1.73\)
- B \(3.0\)
- C \(3.46\)
- D \(6.0\)
Answer & Solution
Correct Answer
(B) \(3.0\)
Step-by-step Solution
Detailed explanation
\begin{array}{l} Given\,:\,m = 0.160kg\\ \theta = {60^ \circ }\\ V = 10\,m/s\\ Angular\,momentu\,m\,L = \vec r \times m\,\vec V\\ = H\,mv\,\cos \theta \\ = \,mv\frac{{{V^2}{{\sin }^2}\theta }}{{2g}}\cos \theta \,\,\,\left[ {H = \frac{{{V^2}{{\sin }^2}\theta }}{{2g}}} \right]\\ =…
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