Question

A bullet of mass $50g$ is projected upward with an initial velocity $200m{s}^{-1}$ in a direction of angle of projection ${60}^o$. The angular momentum of bullet when it is crossing the highest point with respect to point of projection is : $(g=10m{s}^{-2})$

• A. $7.5\times {10}^{2}kg{m}^2{s}^{-1}$
• B. $7.5\times {10}^{3}kg{m}^3{s}^{-1}$
• C. $5\times {10}^{2}kg{m}^2{s}^{-1}$
• D. $5\times {10}^{3}kg{m}^3{s}^{-1}$

Answer 1

The correct option is B $7.5\times {10}^{3}kg{m}^3{s}^{-1}$

When the bullet reaches the highest point, the vertical component of velocity will be zero,

Velocity at the highest point will be $V=200cos60=100m/s$

Maximum height $h=\frac{(200sin60{)}^2}{2g}=1500$

So at highest point, angular momentum $=mVh=50\times {10}^{-3}\times 100\times 1500=7.5\times {10}^{3}kg{m}^3/s$