In basketball, hang is an illusion in which a player seems to weaken the gravita

Question

In basketball, hang is an illusion in which a player seems to weaken the gravitational acceleration while in midair. The illusion depends much on a skilled player's ability to rapidly shift the ball between hands during the flight, but might also be supported by the longer horizontal distance the player travels in the upper part of the jump than in the lower part. If a player jumps with an initial speed of v0 = 7.20 m/s at an angle of ?0 = 41.5°, what percent of the jump's range does the player spend in the upper half of the jump (between maximum height and half maximum height)?

Explanation / Answer

initial velocity = v, angle = A maximum height = h = [vsin(A)]^2/(2g) time from starting to the maximum height = T = vsin(A)/g lower part: vertical displacement = h/2, vertical initial velocity = vsin(A), acceleration a = -g time = t h/2 = vsin(A)*t - gt^2/2 gt^2/2 - vsin(A)*t + h/2 = 0 solve it, use h = [vsin(A)]^2/(2g) and get t = T[1 - 1/sqrt(2)] range = vcos(A)*t upper part: time = T - t range = vcos(A)*(T - t) the required percentage = [vcos(A)*(T - t)]/[vcos(A)*T] = 1 - t/T = 1/sqrt(2) = 71%

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