Final answer:
The average velocity of a projectile at the same height at two different times is its constant horizontal velocity component, which is u cos θ.
"the correct option is approximately option C"
Step-by-step explanation:
Average Velocity of a Projectile in Motion
When a particle is projected with speed u at an angle θ with the horizontal from the ground, and it reaches the same height at times t₁ and t₂, the average velocity during this time interval can be calculated considering the horizontal displacement. The vertical motion does not affect the average velocity since the particle returns to the same height, meaning the vertical displacement is zero.
The horizontal velocity component, u cos θ, remains constant throughout the motion because the acceleration in the horizontal direction, aₓ, is zero. Since the projectile is at the same height at t₁ and t₂, it has covered a horizontal distance of u cos θ × (t₂ - t₁).
The average velocity is the total displacement divided by the elapsed time. As the vertical displacement is zero, the horizontal displacement defines the average velocity. Therefore, the average velocity in the time interval t₁ to t₂ is u cos θ.