a particle of mass m is projected with velocity v making an angle of 45° with the horizontal The magnitude of the angular momentum of the projectile about the point of projection when the particle is at the maximum height h is
a)0 b)mv3 /4√2 g c)mv3 /√2 g d)m√(2gh3)
I am getting the ans as mv3 /2√2 g
please give me the correct solution
1yes the ans is b....
the velocity at maximum height is v/√2...let it be v0(becoz horizontal velocity is same at every point in the motion and is given by ucosθ)
maximum height is v2/4g....which is let r(by the formula u2sin2θ/2g....)
THerefore, angular momentum is mv0r...
here,θ=45°
11The ans is :- B) mv3 / (4g√2)
When the particle is at top the velocity is v cos 45 since the velocity is only having an x-component and not a y-component
V= vcos45
The time required to reach the top is vsin45/g
There The height = r = vsin45 * vsin45/g - 1/2* g*v2sin245/g2
On solving
r = v2sin245/2g
The angular momentum is mvr
m*vcos45*v2sin245/2g
m*v(1/√2)*v2(1/√2)2/(2g)
Therefore
The angular momentum is mv3/(4g√2)
