A particle is projected. If after 2 , s it ma | vedclass

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(B) Let the initial velocity be $u$ and the angle of projection be $	heta$.At the highest point,the vertical component of velocity is zero. The parti

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

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(B) Let the initial velocity be $u$ and the angle of projection be $	heta$.At the highest point,the vertical component of velocity is zero. The particle reaches the highest point $1 \, s$ after the $2 \, s$ mark,so the total time to reach the highest point is $T = 2 + 1 = 3 \, s$.Using $T = \frac{u \sin 	heta}{g} = 3 \, s$,we get $u \sin 	heta = 3g = 30 \, m/s$ ... $(i)$At $t = 2 \, s$,the velocity makes an angle of $30^\circ$ with the horizontal. The horizontal component of velocity remains constant: $v_x = u \cos 	heta$.The vertical component at $t = 2 \, s$ is $v_y = u \sin 	heta - g(2) = 30 - 20 = 10 \, m/s$.Since $	an 30^\circ = \frac{v_y}{v_x}$,we have $\frac{1}{\sqrt{3}} = \frac{10}{u \cos 	heta}$,so $u \cos 	heta = 10\sqrt{3} \, m/s$ ... (ii)Dividing $(i)$ by (ii): $	an 	heta = \frac{30}{10\sqrt{3}} = \sqrt{3}$,so $	heta = 60^\circ$.Substituting $	heta = 60^\circ$ in $(i)$: $u \sin 60^\circ = 30 \implies u(\frac{\sqrt{3}}{2}) = 30 \implies u = \frac{60}{\sqrt{3}} = 20\sqrt{3} \, m/s$.

$A$ particle is projected. If after $2 \, s$ it makes an angle of $30^\circ$ with the horizontal and after $1 \, s$ more it is moving horizontally,find the initial velocity and angle of projection of the particle.

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