Question

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 θ with the horizontal.

Answer 1

y = u sin θ t - 1/2 g t²               vy = u sinθ - g t
x = u cos θ t                            vx = u cosθ

vy = 0 at maximum height,  at  t = T = u sinθ / g
maximum height : H = u² Sin²θ /2g

The path of the projectile is symmetric about the maximum height position.  So the time required to reach from H/2 to H and the time it takes to reach from H again to H/2 are same.

 time to drop a height of H/2 from H with an initial vertical velocity of 0.
               u² Sin²θ/4g = 1/2 g t^2            t = u sinθ / (g√2)

So we need to find the average speed from t = t1 to t2
     t1 = u sinθ/2g * (2-√2)             t2 = u sinθ/2g (2 + √2).

speed v = √[vx² + vy²] = g * √[ (t - u sinθ/g)² + u² cosθ²/g² ]

let      w = (t - u sinθ/g) / (u cosθ/g)
         dw = g dt / (ucosθ) 
 Limits of integration:   w1 = -1/√2            w2 = 1/√2

v = u cosθ √(1+ w²)

$Average\ speed\ v_{avg}= \frac{1}{t2-t1}\int\limits^{t2}_{t1} {v} \, dt\\\\ =\frac{1}{(u\ cos\theta\sqrt2/g)}\frac{u\ cos\theta}{g}\int\limits^{\frac{1}{\sqrt2}}_{-\frac{1}{\sqrt2}} {u\ cos\theta\sqrt{1+w^2}} \, dw\\\\ =\frac{u\ cos\theta}{\sqrt2}\ Tan^{-1}[ w ]^\frac{1}{\sqrt2}_\frac{-1}{\sqrt2}\\\\=\sqrt2\ u\ cos\theta\ Tan^{-1}[ \frac{1}{\sqrt2}]$

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If we are looking for the average velocity vector v = vx i + vy j  then, the vertical component vy is positive before reaching the peak, and is negative afterwards symmetrically.  Hence the vertical component cancels out.

Average velocity vector $\vec{v_{avg}}=\vec{v_x}=u\ cos\theta\ \hat{i}$