A particle is thrown with a velocity of u , m | vedclass

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(C) The vertical displacement $y$ of a projectile at any time $t$ is given by $y = (u \sin 	heta)t - \frac{1}{2}gt^2$.Since the particle is at the sa

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$40$

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(C) The vertical displacement $y$ of a projectile at any time $t$ is given by $y = (u \sin 	heta)t - \frac{1}{2}gt^2$.Since the particle is at the same height at times $t_1$ and $t_2$,we have:$(u \sin 	heta)t_1 - \frac{1}{2}gt_1^2 = (u \sin 	heta)t_2 - \frac{1}{2}gt_2^2$$(u \sin 	heta)(t_2 - t_1) = \frac{1}{2}g(t_2^2 - t_1^2)$$u \sin 	heta = \frac{g(t_1 + t_2)}{2}$Given $t_1 = 1 \, s$,$t_2 = 3 \, s$,$	heta = 30^{\circ}$,and $g = 10 \, m/s^2$:$u \sin 30^{\circ} = \frac{10(1 + 3)}{2}$$u 	imes 0.5 = \frac{10 	imes 4}{2}$$0.5u = 20$$u = 40 \, m/s$.

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