An object of mass 'm' is projected from origin in a vertical xy plane at an angle $45^{\circ}$ with the x-axis with an initial velocity $\mathrm{v}_{0}$. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [g is acceleration due to gravity]

(D) $\frac{\mathrm{mv}_{0}^{3}}{4 \sqrt{2} \mathrm{~g}}$ along negative z-axis

(A) $\frac{\mathrm{mv}_{0}^{3}}{2 \sqrt{2} \mathrm{~g}}$ along negative z-axis

(B) $\frac{m v_{0}^{3}}{2 \sqrt{2} g}$ along positive $z$-axis

(C) $\frac{m v_{0}^{3}}{4 \sqrt{2} g}$ along positive $z$-axis

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Rotational Mechanics

An object of mass 'm' is projected from origin in a vertical xy plane at an angle

4 5^{\circ}

with the x-axis with an initial velocity

v_{0}

. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [g is acceleration due to gravity]

(A) $\frac{mv _{0}^{3}}{2 2 g}$ along negative z-axis
(B) $\frac{m v _{0}^{3}}{2 2 g}$ along positive $z$ -axis
(C) $\frac{m v _{0}^{3}}{4 2 g}$ along positive $z$ -axis
(D) $\frac{mv _{0}^{3}}{4 2 g}$ along negative z-axis✓ Correct

Answer: (D)

Solution

The object is launched at

4 5^{\circ}

with speed

v_{0}

, so

v_{x} = v_{0} cos 4 5^{\circ} = \frac{v _{0}}{2}

and

v_{y} = v_{0} sin 4 5^{\circ} = \frac{v _{0}}{2}

.

At maximum height the vertical velocity vanishes and only

v_{x} = \frac{v _{0}}{2}

survives, directed along the x-axis. The maximum height reached is

H = \frac{( v _{0} sin 4 5 ^{\circ} ) ^{2}}{2 g} = \frac{v _{0}^{2} /2}{2 g} = \frac{v _{0}^{2}}{4 g}

At this instant the position vector from the origin is

r = x \hat{i} + H \hat{j}

and the velocity is

v = v_{x} \hat{i}

. The angular momentum is

L = m (r \times v) = m (x \hat{i} + H \hat{j}) \times (v_{x} \hat{i}) = - m H v_{x} \hat{k}

L

points along the negative z-axis, with magnitude

L = m H v_{x} = m \cdot \frac{v _{0}^{2}}{4 g} \cdot \frac{v _{0}}{2} = \frac{m v _{0}^{3}}{4 2 g}

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Rotational Mechanics

Solution

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