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(C) The vertical displacement of the projectile between times $t_1$ and $t_2$ is zero because the particle is at the same height at both times.Average

Vedclass · Accepted Answer

$u \cos 	heta$

Vedclass · Answer

(C) The vertical displacement of the projectile between times $t_1$ and $t_2$ is zero because the particle is at the same height at both times.Average velocity is defined as the total displacement divided by the total time interval.Average velocity $\vec{v}_{avg} = \frac{\Delta \vec{r}}{\Delta t} = \frac{\Delta x \hat{i} + \Delta y \hat{j}}{t_2 - t_1}$.Since the height is the same,$\Delta y = 0$.The horizontal displacement $\Delta x$ is given by $v_x 	imes (t_2 - t_1)$,where $v_x = u \cos 	heta$.Thus,$\Delta x = (u \cos 	heta)(t_2 - t_1)$.Substituting this into the average velocity formula: $\vec{v}_{avg} = \frac{(u \cos 	heta)(t_2 - t_1) \hat{i} + 0 \hat{j}}{t_2 - t_1} = u \cos 	heta \hat{i}$.Therefore,the magnitude of the average velocity is $u \cos 	heta$.

$A$ particle is projected with speed $u$ at an angle $\theta$ with the horizontal from the ground. If it is at the same height from the ground at times $t_1$ and $t_2$,then its average velocity in the time interval $t_1$ to $t_2$ is .........

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