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(C) The horizontal component of velocity remains constant throughout the motion of the projectile.Let $u = 147 \ m/s$ and $	heta_1 = 60^{\circ}$. The

Vedclass · Accepted Answer

$5.49$

Vedclass · Answer

(C) The horizontal component of velocity remains constant throughout the motion of the projectile.Let $u = 147 \ m/s$ and $	heta_1 = 60^{\circ}$. The horizontal component is $u_x = u \cos 60^{\circ} = 147 	imes 0.5 = 73.5 \ m/s$.At time $t$,the velocity is $v$ and the angle is $	heta_2 = 45^{\circ}$. The horizontal component is $v_x = v \cos 45^{\circ} = v / \sqrt{2}$.Since $u_x = v_x$,we have $v \cos 45^{\circ} = u \cos 60^{\circ} \Rightarrow v = u \frac{\cos 60^{\circ}}{\cos 45^{\circ}} = 147 	imes \frac{0.5}{1/\sqrt{2}} = 73.5 \sqrt{2} \ m/s$.The vertical component of velocity at time $t$ is $v_y = v \sin 45^{\circ} = (73.5 \sqrt{2}) 	imes (1/\sqrt{2}) = 73.5 \ m/s$.The initial vertical component is $u_y = u \sin 60^{\circ} = 147 	imes \frac{\sqrt{3}}{2} \approx 127.3 \ m/s$.Using the equation $v_y = u_y - gt$,we get $73.5 = 127.3 - 9.8t$.$9.8t = 127.3 - 73.5 = 53.8$.$t = 53.8 / 9.8 \approx 5.49 \ s$.

$A$ projectile is thrown in the upward direction making an angle of $60^{\circ}$ with the horizontal direction with a velocity of $147 \ m/s$. Then the time after which its inclination with the horizontal is $45^{\circ}$ is ......... $s$.

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