InterviewSolution
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InterviewSolution
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A projectile of mass `m` is launched with an initial velocity `vec v_i` making an angle `theta` with the horizontal as shown in figure. The projectile momentum of the particle about the origin when it is at the highest point of its trajectory is. .A. `(-m_i^3 sin theta^2 cos theta)/(2 g) hat k`B. `(m_i^3 sin theta^2 cos theta)/(2 g) hat k`C. `(-m_i^3 sin theta^2 cos theta)/(g) hat k`D. `(2 m_i^3 sin theta^2 cos theta)/(g) hat k` |
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Answer» Correct Answer - A (a) At the highest point of the trajectory, `x = (1)/(2) R = (v_i^2 sin 2 theta)/(2 g)` and `y = h_(max) = ((v_i sin theta)^2)/(2 g)` The angular momentum is then `vec L_1 = vec r_1 xx m vecv_1` =`[(v_i^2 sin 2 theta)/(2 g)hati +((v_i sin theta)^2)/(2g)hat j]xx mv_(x i) hat i` =`(-mv_i^2 sin theta^2 cos theta)/(2 g) hat k`. |
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