An object of mass ' m ' is projected from origin in a vertical xy… | JEE Main 2025 (24 Jan, Shift 1)

Q7JEE Main 2025 · 24 JAN · SHIFT 1Medium

Velocity, Momentum & Energy in Flight

An object of mass ' m ' is projected from origin in a vertical xy plane at an angle $45^\circ$ with the $x\text{-}$ 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]

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Answer: (d)

At the top, $\vec r = x\hat i + H\hat j$ and $\vec p = m v_0\cos 45^\circ\,\hat i$ , so $|\vec L| = m(v_0\cos 45^\circ)H$ .
$H = \frac{v_0^2\sin^2 45^\circ}{2g} = \frac{v_0^2}{4g}$
$L = m \cdot \frac{v_0}{\sqrt{2}} \cdot \frac{v_0^2}{4g} = \frac{m v_0^3}{4\sqrt{2}g}$
$\vec L = (H\hat j) \times (p\,\hat i) = -Hp\,\hat k$ , i.e. along negative $z$ -axis.
So the answer is (d).

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