Question

A basketball star covers 2.65 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 mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.95 m above the floor and is at elevation 0.910 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. m/s (c) Determine his vertical velocity at the instant of takeoff. m/s (d) Determine his takeoff angle above the horizontal (e) For comparison, determine the hang time of a whitetail deer making a jump with center-of-mass elevations yi = 1.20 m, Ymax -2.50 m, and Yf -0.690 m

Answer 1

(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 - 0.910)m / 9.8m/s²) = 0.4607 s
so his total hang time is t + t' = 0.8963 s

(b) Vx = x / t = 2.65m / 0.8963s = 2.956 m/s

(c) Use y = Yo + Vo*t + ½at²
0.910 m = 1.02m + Vy*0.8963s - 4.9m/s²*(0.873s)²
Vy = 4.0432 m/s

(d) Θ = arctan(Vy/Vx) = 53.829º

(e) rising: t = √(2h / g) = √(2*(2.50 - 1.20)m / 9.8m/s²) = 0.515 s
falling: t' = √(2*(2.50 - 0.690)m / 9.8m/s²) = 0.6077 s
so his total hang time is t + t' = 1.1227 s