Question

A particle is projected with a speed $v$ at ${45}^{\circ }$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

• A. Zero
• B. $\frac{mv{h}^2}{\sqrt{2}}$
• C. $\frac{m{v}^{2}h}{2}$
• D. $\frac{mv{h}^3}{\sqrt{3}}$
• E. $\frac{mvh}{\sqrt{2}}$

Answer 1

The correct option is E $\frac{mvh}{\sqrt{2}}$

In projectile motion, angular momentum $\left(w\right)$ is given as:

$w=mu\cos \theta h$

where, $w=$ angular momentum $\theta =$ angle of projection

$u=$ velocity of the object at projection

$h=$height

Here,

$u=v$

$\theta ={45}^o$

$h=n$

$w=mv\cos {45}^o\times h$

$w=\frac{mvh}{\sqrt{2}}$

$\therefore E$ is the correct option.