Question

A projectile of mass m is launched with an initial velocity $\overrightarrow{v_1}$ making an angle $\theta$ with the horizontal as shown in figure. The projectile moves in the gravitational field to the Earth. The angular momentum of the projectile about the origin when it is at the highest point of its trajectory is:

• A. $\frac{-m{v}_i^3{\sin }^2\theta \cos \theta }{2g}\hat{k}$
• B. $\frac{m{v}_i^3{\sin }^2\theta \cos \theta }{2g}\hat{k}$
• C. $\frac{-m{v}_i^3{\sin }^2\theta \cos \theta }{g}\hat{k}$
• D. $\frac{2m{v}_i^3{\sin }^2\theta \cos \theta }{g}\hat{k}$

Answer 1

Answer: (A)

$\frac{-m{v}_i^3{\sin }^2\theta \cos \theta }{2g}\hat{k}$

Angular momentum =$m\overrightarrow{r}\times \overrightarrow{v}$

At the highest point $x=\frac{1}{2}R=\frac{u^{2}sin2\theta }{2g}$

$y=\frac{u^{2}si{n}^2\theta }{2g}$

Angular momentum= $L=[\frac{u^{2}sin2\theta }{2g}\hat{i}+\frac{u^{2}si{n}^2\theta }{2g}\hat{j}]\times mucos\theta \hat{i}$

gives, $L=\frac{m{u}^{3}si{n}^2\theta cos\theta }{2g}(-\hat{k})$