Question

average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory is speed of projection is u angle of projection from horizontal equal to theta​

Answer 1

Answer:

u/2 ( 1+ cos∝)

Explanation:

let theta = ∝

at the highest point : vx = ucos∝ , vy = 0 , v = √vx² + vy² = ucos∝

at the lowest point :  vx = ucos∝ , vy = u sin∝ , v = u

average speed = u +  ucos∝ ÷2 = u/2 ( 1+ cos∝)