A particle is projected with a speed v at 45° 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

Question

A particle is projected with a speed v at 45° 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) (mvh2/√2) (C) (mv2 h/2) (D) (mvh/√2)

Tardigrade Admin · Accepted Answer

When a particle is projected with a speed v at 45

^{\circ}

with the horizontal then velocity of the projectile at maximum height

v^{'} = v cos 4 5^{\circ}

=

\frac{v}{2}

Angular momentum of the projectile about the point of projection = mv' h

= m \frac{m}{2} h = \frac{m v h}{2}

Q. 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

zero

$\frac{m v h ^{2}}{2}$

$\frac{m v ^{2} h}{2}$

$\frac{m v h}{2}$

When a particle is projected with a speed v at 45 $^{\circ}$ with the horizontal then velocity of the projectile at maximum height
$v^{'} = v cos 4 5^{\circ}$ = $\frac{v}{2}$
Angular momentum of the projectile about the point of projection = mv' h
$= m \frac{m}{2} h = \frac{m v h}{2}$

Q. A particle is projected with a speed v at 45∘ 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

zero

2​mvh2​

2mv2h​

2​mvh​

When a particle is projected with a speed v at 45∘ with the horizontal then velocity of the projectile at maximum height v′=vcos45∘=2​v​ Angular momentum of the projectile about the point of projection = mv' h =m2​m​h=2​mvh​

Q. 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

$\frac{m v h ^{2}}{2}$

$\frac{m v ^{2} h}{2}$

$\frac{m v h}{2}$

When a particle is projected with a speed v at 45 $^{\circ}$ with the horizontal then velocity of the projectile at maximum height
$v^{'} = v cos 4 5^{\circ}$ = $\frac{v}{2}$
Angular momentum of the projectile about the point of projection = mv' h
$= m \frac{m}{2} h = \frac{m v h}{2}$