A ball of mass 160g is thrown up at an angle of 60° to the horizontal at a speed of 10ms⁻¹. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly: (g=10ms⁻²) 1.73 kgm²/s, 3.0 kgm²/s, 6.0 kgm²/s, 3.46 kgm²/s

The initial components of velocity are,
V =5i+5√3j
At the highest point, only the horizontal component will remain.
Hence,v=5i
The vector corresponding to the highest point is given by,
T =2usinθ/g=2(5√3)/10=√3
x =5√3
y=ut -1/2gt²
y=0 at highest point
Therefore, time taken to reach highest point t= √3
x =5t=5√3
y=0
Hence, the angular momentum is given by the cross product of the radius vector with the velocity vector.
L = m (r⃗ × v⃗)=0.16(5√3i × 5i) = 0.16(0) = 0
The horizontal distance at the highest point is given by:
Range = u²sin2θ/g
R = (10)² sin(260°)/10 = 10√3/2 = 5√3 m
Height H = u²sin²θ/2g
H = 10² sin²(60°)/(210) = 100(3/4)/(20) = 3.75 m
At the highest point, the velocity of the ball is horizontal and is given by:
vₓ = v₀cosθ = 10cos60° = 5 m/s
The position vector of the ball at the highest point with respect to the point of projection is:
r = R/2 î + Hĵ = (5√3/2)î + 3.75ĵ
The angular momentum L is given by:
L = m(r x v) = 0.16[(5√3/2)î + 3.75ĵ] x 5î
Since î x î = 0, we have:
L = 0.16(3.75ĵ x 5î) = 0.16(3.75 * 5)(-k) = -3 kg m²/s
Magnitude of angular momentum is 3 kg m²/s