A particle is projected vertically upward with speed u and returns to the point of projection after total time T. Its average speed during the first 3T/4 is:
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A particle is projected vertically upward with speed u and returns to the point of projection after total time T. Its average speed during the first 3T/4 is:

G1 Physics · Accepted Answer

Since the complete return time is T = 2u/g, we have 3T/4 = 3u/(2g). The particle reaches the highest point at u/g. Distance upward: s₁ = u²/(2g). During the next time u/(2g), it falls from rest through s₂ = (g/2)[u/(2g)]² = u²/(8g). Total distance in the first 3T/4: s = 5u²/(8g). Average speed: s / [3u/(2g)] = 5u/12.