Question

Average velocity of a particle in projectile between its starting point and the highest point of its trajectory is(projection speed=u,Angle of projection from horizontal=theta,)

Answer 1

Average magnitude of the velocity is required ?

Vx = u CosФ ,  Vy = u SinФ  - g t
V = √[(u CosФ)² + (u Sin Ф - g t)² ]
   = u CosФ √[ 1 + (tan Ф - g secФ/u * t)² ]

Let z = TanФ - (g SecФ /u) t ;    dt = - (u CosФ/g) dz 

V = u CosФ √[ 1 + z² ]

To find average of V over time t = 0 sec, to T = u SinФ /g   ie., till the highest point.
When t =0,  z = TanФ.     When t = T,  z = 0

V_avg = 1/T * Integral from t=0 to  T,   { V } dt
           = 1/T * u² Cos² Ф /g * Integral from z = 0 to TanФ, { √(1+z²) } dz
           = (u Cos Ф Cot Ф) * Int from z = 0 to TanФ , { √(1+z²) } dz 

 Let z = Tan w.    dz = Sec² w dw,
    When  z = 0,  w = 0.  And,  when z = TanФ,  w = Ф.
V_avg = (u Cos Ф * Cot Ф) * Integral w = 0 to Ф,  { Sec³ w } dw
     = (u CosФ *CotФ) * 1/2 [ Sec w tan w + Ln (SecФ+tanФ) ]_0 to Ф
     = (u CosФ *CotФ)/2 * [ Sec Ф TanФ + Ln (secФ+ tanФ) ]

For example if  Ф = 45°,  
V_avg = u / (2√2) * [√2 + Ln (√2 +1) ]