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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 $v_y = u \sin 	heta$.At $t = 3 \, s$,the projectile is at its highest point,so $v_y = 0$. Thus,$u \sin 	heta - g(3) = 0$,which gives $u \sin 	heta = 3g = 30 \, m/s$ (taking $g = 10 \, m/s^2$).At $t = 2 \, s$,the vertical velocity is $v_y' = u \sin 	heta - g(2) = 30 - 20 = 10 \, m/s$.The horizontal velocity remains constant: $v_x = u \cos 	heta$.The direction is given by $	an 30^o = \frac{v_y'}{v_x} = \frac{10}{u \cos 	heta}$.Since $	an 30^o = \frac{1}{\sqrt{3}}$,we have $\frac{1}{\sqrt{3}} = \frac{10}{u \cos 	heta}$,so $u \cos 	heta = 10\sqrt{3} \, m/s$.The initial velocity magnitude is $u = \sqrt{(u \cos 	heta)^2 + (u \sin 	heta)^2} = \sqrt{(10\sqrt{3})^2 + (30)^2} = \sqrt{300 + 900} = \sqrt{1200} = 20\sqrt{3} \, m/s$.The angle $	heta$ is $	an 	heta = \frac{u \sin 	heta}{u \cos 	heta} = \frac{30}{10\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$. Thus,$	heta = 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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