Question

A body of mass (2)^1/2 kg is projected with a speed V at an angle 45° to the horizontal, When the body
is at the highest point in it's path, angular momentum of the body about the point of projection is
(i,j are unit vectors in vertical plane)​

Answer 1

Answer:

$\frac{v^{3} }{4g}$

Explanation:

When the velocity of the body is horizontal,

angular momentum about the point of projection = mass*velocity * max.height

                                                                                  = $\sqrt{2}$*$\frac{v}{\sqrt{2} }$ *$\frac{v^{2}( \frac{1}{2}) }{2g}$

                                                                                  = $\frac{v^{3} }{4g}$.

Answer 2

Concept introduction:

At highest point of of projectile only horizontal velocity acts.

Explanation:

Given that, a body of mass $2^{1/2} kg$ is projected with a speed V at an angle $45$° to the horizontal

We have to find, angular momentum of the body at highest point of the projectile.

According to the question,

When the velocity of the body is horizontal,

Angular momentum about the point of projection = mass*velocity * max. height

= $\sqrt{2} *\frac{V}{\sqrt{2} } *\frac{V^{2} }{4g} =\frac{V^{3} }{4g}$

Final Answer:

Angular momentum about the point of projection =$\frac{V^{3} }{4g}$.

#SPJ3