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(A) The projectile is projected with speed $u$ at an angle $	heta$ with the horizontal.Let the projectile cross the same level at times $t_1$ and $t_

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$u \cos 	heta$

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(A) The projectile is projected with speed $u$ at an angle $	heta$ with the horizontal.Let the projectile cross the same level at times $t_1$ and $t_2$.The horizontal component of velocity is $v_x = u \cos 	heta$,which remains constant throughout the motion.The vertical component of velocity is $v_y = u \sin 	heta - gt$.At the same level,the vertical displacement is zero,so the vertical velocity components at $t_1$ and $t_2$ are equal in magnitude but opposite in direction.The horizontal displacement between these two instants is $\Delta x = (u \cos 	heta) 	imes \Delta t$,where $\Delta t = t_2 - t_1$.The average velocity is defined as $\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t}$.Since the vertical displacement is zero,the average velocity is purely horizontal: $\vec{v}_{avg} = \frac{\Delta x}{\Delta t} = \frac{(u \cos 	heta) \Delta t}{\Delta t} = u \cos 	heta$.

$A$ projectile is projected with speed $u$ at an angle $\theta$ with the horizontal. The average velocity of the projectile between the instants it crosses the same level is ............

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