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(C) The horizontal component of velocity remains constant throughout the motion.$u_x = u \cos 60^{\circ} = 150 	imes \frac{1}{2} = 75\, ms^{-1}$.Let

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$7.5(\sqrt{3} - 1)\,s$

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(C) The horizontal component of velocity remains constant throughout the motion.$u_x = u \cos 60^{\circ} = 150 	imes \frac{1}{2} = 75\, ms^{-1}$.Let the velocity at time $t$ be $v$. At this time,the angle with the horizontal is $45^{\circ}$.The horizontal component of velocity at time $t$ is $v_x = v \cos 45^{\circ}$.Since $v_x = u_x$,we have $v \cos 45^{\circ} = 75$,so $v = \frac{75}{\cos 45^{\circ}} = 75\sqrt{2}\, ms^{-1}$.The vertical component of velocity at time $t$ is $v_y = v \sin 45^{\circ} = 75\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 75\, ms^{-1}$.Using the equation $v_y = u_y - gt$,where $u_y = u \sin 60^{\circ} = 150 	imes \frac{\sqrt{3}}{2} = 75\sqrt{3}\, ms^{-1}$:$75 = 75\sqrt{3} - 10t$.$10t = 75(\sqrt{3} - 1)$.$t = 7.5(\sqrt{3} - 1)\,s$.

$A$ projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $150\, ms^{-1}$. Then the time after which its inclination with the horizontal is $45^{\circ}$ is

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