Question
A particle is projected from the ground with an initial speed $\upsilon $ at an angle $\theta $ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is
✓
Answer
Average velocity $=\frac{\text { Displacement }}{\text { Time }}$
$v_{w_{w}}=\frac{\sqrt{H^{2}+\frac{R^{2}}{4}}}{T / 2}$
Here, $\quad \mathrm{H}=$ maximum height $=\frac{v^{2} \sin ^{2} \theta}{2 g}$
$\mathrm{R}=\mathrm{range}=\frac{\mathrm{v}^{2} \sin 2 \theta}{\mathrm{g}}$
and $\quad \mathrm{T}=$ time of flight $=\frac{2v \sin \theta}{\mathrm{g}}$
$\therefore \quad \mathrm{U}_{\mathrm{av} .}=\frac{\mathrm{v}}{2} \sqrt{1+3 \cos ^{2} \theta}$
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