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Q.

A projectile is fire with velocity 54 km/hr making angle $45^{o}$ with horizontal. Angular momentum of this particle of mass 1 kg about the point of projection after one second into the motion will be $\frac{5 N}{\sqrt{2}}$ in SI unit ( $g = 10 m / s^{2}$ ). Find the value of N

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answer is 15.

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Detailed Solution

u = 54 km/hr = 15 m/sec
torque at time t is $τ =$ $τ = m g u \cos θ t$
$\frac{d l}{d t} = τ$
$\int_{1}^{L} d L = \int_{0}^{1} m g u \cos θ d t$
$L = \frac{m g u \cos θ}{2} = \frac{10 \times 15}{2 \sqrt{2}} = \frac{75}{2} k g m^{2} / \sec$
So, N = 15

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