Question

A particle of mass $m$ is projected with velocity $\vartheta$ making an angle of ${45}^{\circ }$ with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where g=acceleration due to gravity)

• A. $\frac{mv\times v^2}{\sqrt[2]{22}\times 4g}$
• B. $\frac{m{v}^3}{\left(\sqrt{2g}\right)}$
• C. $\frac{m{v}^3}{\left(\sqrt{4}2g\right)}$
• D. $\frac{m{v}^2}{2g}$

Answer 1

The correct option is A $\frac{mv\times v^2}{\sqrt[2]{22}\times 4g}$

So horizontal velocity which remains $constant$ is $vcos{45}^0=\frac{v}{\sqrt[2]{22}}$

$\text{At highest point of the motion its vertical velocity will be zero and it will possess only horizontal velocity}$

$\text{and the perpendicular distance of velocity vector from the starting point will be equal to the height at}$

$\text{that time i.e. maximum height H}$

So $r=H=\frac{v^{2}si{n}^2{45}^0}{2g}=\frac{v^2}{4g}$

So the angular momentum is $p=mvcos{45}^{0}r=\frac{mv\times v^2}{\sqrt[2]{22}\times 4g}$