A particle of mass $m$ is projected with a velocity $v$ making an angle of $45^{o}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at maximum height $h$ is:

Z e r o

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{m v^{3}}{4 \sqrt{2} g}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{m v^{3}}{\sqrt{2} g}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m \sqrt{2 g} h^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{m v^{3}}{4 \sqrt{2} g}$
Velocity at maximum height $V = v c o s 45^{o}$
Maximum height, $h = \frac{v^{2} s i n^{2} 45}{2 g} = \frac{v^{2}}{4 g}$
Angular momentum at point $h$ , $= m V h = \frac{m v^{3}}{4 \sqrt{2} g}$

Suggest Corrections

Similar questions

Q. A particle of mass

m

is projected with a velocity

v

making an angle of

45^{\circ}

with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height

h

is given by which of the following relation(s)?

A particle of mass m is projected with velocity v 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)

Q. A particle of mass m is projected with a velocity v making an angle of 30 with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height h is

Q. A particle of mass

m

is projected with a velocity

V

making an angle of

30^{\circ}

with the horizontal. The magnitude of the angular momentum of the projectile motion about the point of projection when the particle is at maximum height

H

is given by which of the following expression(s)?

A particle of mass m is projected with velocity v making an angle of $45^{\circ}$ with the horizontal. Themagnitude 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)

Join BYJU'S Learning Program

A particle of mass m is projected with a velocity v making an angle of 45o with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at maximum height h is:

The correct option is C mv34√2gVelocity at maximum height V=v cos 45oMaximum height, h=v2sin2452g=v24gAngular momentum at point h, =mVh=mv34√2g

A particle of mass $m$ is projected with a velocity $v$ making an angle of $45^{o}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at maximum height $h$ is:

The correct option is C $\frac{m v^{3}}{4 \sqrt{2} g}$
Velocity at maximum height $V = v c o s 45^{o}$
Maximum height, $h = \frac{v^{2} s i n^{2} 45}{2 g} = \frac{v^{2}}{4 g}$
Angular momentum at point $h$ , $= m V h = \frac{m v^{3}}{4 \sqrt{2} g}$