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A particle is projected from the ground with an initial speed of $v$ at an angle of projection $\theta$. The average velocity of the particle between its time of projection and time it reaches highest point of trajectory is
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Verified Answer
The correct answer is:
$\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}$
$\begin{aligned} & \text { We know, average velocity }=\frac{\text { displacement }}{\text { time }} \\ & \qquad v_{\text {av }}=\frac{\sqrt{\mu^2+R^2 / 4}}{T / 2}\end{aligned}$
where, $H=$ maximum height $=\frac{v^2 \sin ^2 \theta}{2 g}$
Range $R=\frac{v^2 \sin 2 \theta}{g}$
Time of flight $T=\frac{2 v \sin \theta}{g}$
Putting the values of Eqs. (ii), (iii) and (iv) in Eq. (i) we have
$$
v_{\mathrm{av}}=\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}
$$
where, $H=$ maximum height $=\frac{v^2 \sin ^2 \theta}{2 g}$
Range $R=\frac{v^2 \sin 2 \theta}{g}$
Time of flight $T=\frac{2 v \sin \theta}{g}$
Putting the values of Eqs. (ii), (iii) and (iv) in Eq. (i) we have
$$
v_{\mathrm{av}}=\frac{v}{2} \sqrt{1+3 \cos ^2 \theta}
$$
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