A basketball star covers 2.80 m horizontally in a Jump to dunk the ball. His mot

Question

A basketball star covers 2.80 m horizontally in a Jump to dunk the ball. His motion through space can be modeled precisely as that of a particle at his center of mass. His center of mess is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.9S m above the floor and is at elevation 0.870 m when he touches down again. (a) Determine his time of flight (his "hang time'). (b) Determine his horizontal velocity at the instant of takeoff. (c) Determine his vertical velocity at the instant of takeoff. (d) Determine his takeoff angle. (e) For comparison, determine the hang time of a whitetall deer making a Jump with center-of-mass elevations y_1 = 1.20 m, y_max = 2.25 m, and y_f = 0.680 m.

Explanation / Answer

(a) Instead of having him rise from 1.02 m to 1.95 m, let's imagine him falling instead. The time is the same.
rising: t = (2h / g) = (2*(1.95 - 1.02)m / 9.8m/s²) = 0.4356 s
falling: t' = (2*(1.95 - 0.870)m / 9.8m/s²) = 0.469 s
so his total hang time is t + t' = 0.905 s

(b) Vx = x / t = 2.80m / 0.905s = 3.094 m/s

(c) Use y = Yo + Vo*t + ½at²
0.87 m = 1.02m + Vy*0.905s - 4.9m/s²*(0.905s)²
Vy = 4.27 m/s

(d) = arctan(Vy/Vx) = 54.06º

(e) rising: t = (2h / g) = (2*(2.25 - 1.20)m / 9.8m/s²) = 0.463 s
falling: t' = (2*(2.25 - 0.680)m / 9.8m/s²) = 0.566 s
so his total hang time is t + t' = 1.03 s

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