A particle is projected from the ground with an initial speed of v at an angle of projection q the average velocity of the particle between its time of projection and times it reaches highest point of trajectory is
Correct answer: D. v cos q
- A. v/2√(1+2cos2q)
- B. v/2√(1+2sin2q)
- C. v/2√(1+3cos2q)
- D. v cos q
Explanation
v is the initial speed of the particle.θ is the angle of projection.Here's the explanation:Horizontal and Vertical Components:The initial velocity can be resolved into two components:Horizontal component (v_x): v * cos(θ)Vertical component (v_y): v * sin(θ)Motion at the Highest Point:At the highest point of its trajectory, the vertical component of the particle's velocity becomes zero. However, the horizontal component remains constant throughout the motion due to the absence of a horizontal force.Average Velocity:Since the time spent in the air is symmetrical (equal time taken to reach the highest point and come back down), the average velocity over the entire time interval is equal to the horizontal component of the initial velocity:Average velocity = v * cos(θ)Therefore, the average velocity of the particle between the time of projection and the time it reaches the highest point of its trajectory is v * cos(θ)
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