Solve the following :


Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed $u$ at an angle $\theta$ with the horizontal


Let particle reaches from $A$ to $B$ in time t.

By symmetry, $A B$ line is horizontal

So, displacement $A B=u_{x} t$

Velocity $=\stackrel{\text { displacement }}{\text { time }}=\frac{\text { uxt }}{t}$

$\mathrm{V}_{\mathrm{avg}}=\mathrm{u} \cos \theta$

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