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.
.

Options

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`

Answer

A

Explanation

(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`.