Question

angular momentum of projectile.about point of projection when it strikes the ground is​

Answer 1

To find:

Angular momentum of projectile when it strikes the ground with reference to the point of projection.

Calculation:

At a time t ; velocity will be:

$1) \: v_{x} = u \cos( \theta) \\2) \: v_{y} = u \sin( \theta) - gt$

At time t , position vector will be ;

$1) \: x = u \cos( \theta) t \\ 2) \: y = u \sin( \theta) t - \dfrac{1}{2} g {t}^{2}$

So, angular momentum at time t ;

$\therefore \: \vec{L} = m( \vec{r} \times \vec{v})$

$= > \: \vec{L} = m \bigg \{( x \hat{i} + y \hat{j}) \times ( v_{x} \hat{i} + v_{y} \hat{j}) \bigg \}$

$= > \: \vec{L} = \dfrac{1}{2} mug \cos( \theta) {t}^{2} \: \: ( - \hat{k})$

So putting value of t ;

$= > \: \vec{L} = \dfrac{1}{2} mug \cos( \theta) {( \dfrac{2u \sin( \theta) }{g}) }^{2} \: \: ( - \hat{k})$

$= > \: \vec{L} = \dfrac{2 m {u}^{3}}{g} \cos( \theta) { \sin}^{2} ( \theta) \: \: ( - \hat{k})$

So, final answer is:

$\boxed{ \sf{ \green{ \: \vec{L} = \dfrac{2 m {u}^{3}}{g} \cos( \theta) { \sin}^{2} ( \theta) \: \: ( - \hat{k})}}}$