A particle is projected at \(t\)\(=\)\(0\) from a point \(P\) on the ground with a speed \(v_\circ\) at an angle of \(45^\circ\) to the horizontal. The magnitude of the angular momentum of the particle about \(P\) at time \(t\)\(=\)\((\frac{v_\circ}{g}\)) can be expressed as \(4\)\(\times\)\(10^\tau\)\(.\) Find \(\tau\).

**Note**: Take \(v_\circ\) to be \(1\)\(\frac{m}{s}\), mass of particle to be \(1 \text{ kg}\), \(\sqrt{2}\)\(=\)\(1.4\) and \(g\)\(=\)\(10(\frac{m}{s^2}\))\(.\)

Also make approximations where necessary.

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