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(A) Let the initial velocity be $u$ at an angle $	heta$ with the horizontal. The horizontal component of velocity $u_x = u \cos 	heta$ remains const

Vedclass · Accepted Answer

$40 \sqrt{3} \ ms^{-1}$

Vedclass · Answer

(A) Let the initial velocity be $u$ at an angle $	heta$ with the horizontal. The horizontal component of velocity $u_x = u \cos 	heta$ remains constant throughout the motion.At any time $t$,the vertical component of velocity is $v_y = u \sin 	heta - gt$.The angle $\alpha$ with the horizontal at time $t$ is given by $	an \alpha = \frac{v_y}{u_x} = \frac{u \sin 	heta - gt}{u \cos 	heta}$.At $t_1 = 4 \ s$,$\alpha = 30^{\circ}$:$	an 30^{\circ} = \frac{u \sin 	heta - 10(4)}{u \cos 	heta} \implies \frac{1}{\sqrt{3}} = \frac{u \sin 	heta - 40}{u \cos 	heta} \implies u \cos 	heta = \sqrt{3}(u \sin 	heta - 40) \quad (1)$At $t_2 = 4 + 2 = 6 \ s$,the stone is moving horizontally,so $\alpha = 0^{\circ}$:$	an 0^{\circ} = \frac{u \sin 	heta - 10(6)}{u \cos 	heta} = 0 \implies u \sin 	heta = 60 \ ms^{-1} \quad (2)$Substitute $(2)$ into $(1)$:$u \cos 	heta = \sqrt{3}(60 - 40) = 20 \sqrt{3} \ ms^{-1} \quad (3)$The magnitude of the initial velocity $u$ is $\sqrt{(u \sin 	heta)^2 + (u \cos 	heta)^2} = \sqrt{(60)^2 + (20 \sqrt{3})^2} = \sqrt{3600 + 1200} = \sqrt{4800} = 40 \sqrt{3} \ ms^{-1}$.

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