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)

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!

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

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

Open in App

Solution

The correct option is B $\frac{m v^{3}}{4 \sqrt{2} g}$
$L = \frac{m u^{3} c o s θ s i n^{2} θ}{2 g} = \frac{m v^{3}}{4 \sqrt{2} g} [A s θ = 45^{\circ}]$

Suggest Corrections

Similar questions

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)

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)?

Q. 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)

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 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 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

Join BYJU'S Learning Program

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 particle about the point of projection when the particle is at its maximum height is (where g = acceleration due to gravity)

The correct option is B mv34√2gL=m u3 cos θ sin2 θ2g=mv34√2g [As θ=45∘]

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)

The correct option is B $\frac{m v^{3}}{4 \sqrt{2} g}$
$L = \frac{m u^{3} c o s θ s i n^{2} θ}{2 g} = \frac{m v^{3}}{4 \sqrt{2} g} [A s θ = 45^{\circ}]$