A projectile is fired at 30^{circ} to the hor | vedclass

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(B) Given the vertical component of velocity $u_y = u \sin 	heta = 80 \; ms^{-1}$ and $	heta = 30^{\circ}$.Since $\sin 30^{\circ} = 0.5$,we have $u

Vedclass · Accepted Answer

$80\sqrt{3} \; ms^{-1}$

Vedclass · Answer

(B) Given the vertical component of velocity $u_y = u \sin 	heta = 80 \; ms^{-1}$ and $	heta = 30^{\circ}$.Since $\sin 30^{\circ} = 0.5$,we have $u = \frac{80}{0.5} = 160 \; ms^{-1}$.The horizontal component of velocity is $u_x = u \cos 	heta = 160 \cos 30^{\circ} = 160 	imes \frac{\sqrt{3}}{2} = 80\sqrt{3} \; ms^{-1}$.At time $t = \frac{T}{2}$,the projectile is at its maximum height,where the vertical component of velocity $v_y = 0$.Therefore,the resultant velocity at $t = \frac{T}{2}$ is equal to the horizontal component $u_x$,which remains constant throughout the motion.Thus,$v = u_x = 80\sqrt{3} \; ms^{-1}$.

$A$ projectile is fired at $30^{\circ}$ to the horizontal. The vertical component of its velocity is $80 \; ms^{-1}$. Its time of flight is $T$. What will be the velocity of the projectile at $t = \frac{T}{2}$?

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