A projectile is thrown in the upward directio | vedclass

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(C) Given, angle of projection, $	heta = 60^{\circ}$ and initial velocity, $u = 140 \,ms^{-1}$.The velocity is divided into two components: horizonta

Vedclass · Accepted Answer

$5.124$

Vedclass · Answer

(C) Given, angle of projection, $	heta = 60^{\circ}$ and initial velocity, $u = 140 \,ms^{-1}$.The velocity is divided into two components: horizontal component $u_x = u \cos 60^{\circ} = 140 	imes 0.5 = 70 \,ms^{-1}$ and vertical component $u_y = u \sin 60^{\circ} = 140 	imes \frac{\sqrt{3}}{2} = 70\sqrt{3} \,ms^{-1}$.Let after time $t$, the velocity vector makes an angle of $45^{\circ}$ with the horizontal. At time $t$, the horizontal component $v_x = u_x = 70 \,ms^{-1}$ and the vertical component $v_y = u_y - gt = 70\sqrt{3} - 10t$.Since the angle is $45^{\circ}$, $	an 45^{\circ} = \frac{v_y}{v_x} = 1$, which implies $v_y = v_x$.Substituting the values: $70\sqrt{3} - 10t = 70$.$10t = 70\sqrt{3} - 70$.$t = 7(\sqrt{3} - 1) = 7(1.732 - 1) = 7(0.732) = 5.124 \,s$.

$A$ projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal with a velocity of $140 \,ms^{-1}$. Then the time after which its velocity makes an angle $45^{\circ}$ with the horizontal is (Acceleration due to gravity, $g=10 \,ms^{-2}$) (in $\,s$)

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