An object of mass ' m ' is projected from origin in a vertical xy plane at an angle 45° with the x- axis with an initial velocity v₀. 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]

Solution

Physics sir · 4 steps

At the top the velocity is purely horizontal, so the lever arm is just the maximum height H.

1. State at the top. At maximum height the vertical velocity is zero. Only vₓ = v₀ cos 45° = v₀/√2 is left, along +x. The height is H = (v₀² sin² 45°)/2g = v₀²/4g.
2. Lever arm. For a horizontal velocity, the perpendicular distance from the origin to the line of motion is the height H. The horizontal position x does not matter.

   L = m vₓ H = m · v₀/√2 · v₀²/4g = (m v₀³)/(4√2 g)

3. Direction. L⃗ = m r⃗ × v⃗ with r⃗ = xî + Hĵ and v⃗ = vₓî. The î × î term is zero, and ĵ × î = -k̂, so L⃗ points along -z.
4. So. The angular momentum is (m v₀³)/(4√2 g) along the negative z-axis, which is option (d).

The idea to keep. Angular momentum about a point is momentum times the perpendicular distance to its line of motion. Get the direction from the cross product, and never guess it from the picture.

Answer from the NTA final answer key. Solution by PrepWiser.