A projectile object is thrown in the upward d | vedclass

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Question

(A) The horizontal component of velocity remains constant throughout the projectile motion. Let the initial velocity be $v_0 = 140 \ m/s$ at angle $

Vedclass · Accepted Answer

$\frac{14}{\sqrt{3}} \ s$

Vedclass · Answer

(A) The horizontal component of velocity remains constant throughout the projectile motion. Let the initial velocity be $v_0 = 140 \ m/s$ at angle $	heta_0 = 60^{\circ}$. Let the velocity at time $t$ be $v$ at angle $	heta = 30^{\circ}$.Horizontal component: $v_x = v_0 \cos 60^{\circ} = v \cos 30^{\circ}$.$140 	imes \frac{1}{2} = v 	imes \frac{\sqrt{3}}{2} \implies v = \frac{140}{\sqrt{3}} \ m/s$.Vertical component at time $t$: $v_y = v_0 \sin 60^{\circ} - gt = v \sin 30^{\circ}$.$140 	imes \frac{\sqrt{3}}{2} - 10t = \frac{140}{\sqrt{3}} 	imes \frac{1}{2}$.$70\sqrt{3} - 10t = \frac{70}{\sqrt{3}}$.$10t = 70\sqrt{3} - \frac{70}{\sqrt{3}} = 70 \left( \frac{3-1}{\sqrt{3}} ight) = \frac{140}{\sqrt{3}}$.$t = \frac{14}{\sqrt{3}} \ s$.

$A$ projectile object is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal with velocity of $140 \ m/s$. Then, the time after which its velocity makes an angle $30^{\circ}$ with the horizontal is (Take, $g=10 \ m/s^2$)

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