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(B) Let the initial velocity be $u$ at an angle $	heta$ with the horizontal. The horizontal component is $v_x = u \cos 	heta$ and the vertical compo

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$20\sqrt{3} \, m/s, 60^o$

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(B) Let the initial velocity be $u$ at an angle $	heta$ with the horizontal. The horizontal component is $v_x = u \cos 	heta$ and the vertical component is $v_y(t) = u \sin 	heta - gt$.At $t = 3 \, s$,the projectile is at its highest point,so $v_y(3) = 0$.$u \sin 	heta - g(3) = 0 \implies u \sin 	heta = 3g = 30 \, m/s$ (taking $g = 10 \, m/s^2$).At $t = 2 \, s$,the angle of inclination is $30^o$,so $	an 30^o = \frac{v_y(2)}{v_x}$.$\frac{1}{\sqrt{3}} = \frac{u \sin 	heta - 2g}{u \cos 	heta} = \frac{30 - 20}{u \cos 	heta} = \frac{10}{u \cos 	heta}$.$u \cos 	heta = 10\sqrt{3} \, m/s$.The magnitude of initial velocity $u = \sqrt{(u \cos 	heta)^2 + (u \sin 	heta)^2} = \sqrt{(10\sqrt{3})^2 + 30^2} = \sqrt{300 + 900} = \sqrt{1200} = 20\sqrt{3} \, m/s$.The direction $	heta = 	an^{-1}(\frac{u \sin 	heta}{u \cos 	heta}) = 	an^{-1}(\frac{30}{10\sqrt{3}}) = 	an^{-1}(\sqrt{3}) = 60^o$.

Two seconds after projection,a projectile is travelling in a direction inclined at $30^o$ to the horizontal. After one more second,it is travelling horizontally. What is the magnitude and direction of its velocity at the initial point?

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