Question

A particle of mass m is projected with velocity $v$ making an angle of ${45}^{\circ }$ with the horizontal. If maximum height of the projectile is $h$, then

• A. angular momentum of projectile about origin remains conserved
• B. angular momentum of the projectile about origin when it just starts its motion is zero
• C. angular momentum of projectile about origin when it reaches the highest point is $m\sqrt{g{h}^3}$
• D. angular momentum of projectile about origin when it reaches the highest point is $m\sqrt{2g{h}^3}$.

Answer 1

Answer: (B), (D)

The correct options are
B angular momentum of the projectile about origin when it just starts its motion is zero
D angular momentum of projectile about origin when it reaches the highest point is $m\sqrt{2g{h}^3}$.

About a given point,
angular momentum = moment of linear momentum

Maximum height of Projectile,
$h=\frac{v^2\times si{n}^2{45}^{\circ }}{2g}=\frac{v^2}{4g}$ $\cdots \left(i\right)$

At maximum height, the velocity of projectile
$v\cos \theta =v\cos {45}^{\circ }=\frac{v}{\sqrt{2}}$
$\therefore$ Momentum at highest point $=\frac{mv}{\sqrt{2}}$
$\therefore$ Moment of momentum about point O $=\frac{mv}{\sqrt{2}}\times h$
$\therefore$ Angular momentum about point O $=\frac{mvh}{\sqrt{2}}$
$\therefore L=\frac{mvh}{\sqrt{2}}$ $\cdots \left(ii\right)$

Put $h$ from $\left(i\right)$ in $\left(ii\right)$, we get,
$L=\frac{mv}{\sqrt{2}}\times \left(\frac{v^2}{4g}\right)=\frac{m{v}^3}{4\sqrt{2}g}$ $\cdots \left(iii\right)$

From $\left(i\right)$ and $\left(iii\right)$, eliminating $v$ we get,
Angular momentum $=\frac{m{v}^3}{4\sqrt{2}g}=\frac{m\times (4gh{)}^{3/2}}{4\sqrt{2}g}$
$=m\sqrt{2g{h}^3}$