A basketball star covers 2.60 m horizontally in a jump to dunk the ball (see fig

Question

A basketball star covers 2.60 m horizontally in a jump to dunk the ball (see figure). 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 1.80 m above the floor and is at elevation 0.850 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 whitetail deer making a jump with center-of-mass elevations yi = 1.20 m, ymax = 2.75 m, and yf = 0.800 m.

Explanation / Answer

Rise distance: drise = peak cm height - start cm height
drise = 1/2gt^2
So rise time = sqrt (2 drise / g)
Fall distance: dfall = peak cm height - end cm height
So fall time = sqrt (2 dfall / g)
Total flight time = sqrt (2 (peak height - start height)/g)
+ sqrt (2 (peak height - finish height)/g)

You can then calculate horizontal velocity
vh = distance covered / total flight time

You can calculate initial vertical velocity using conservation of energy on the vertical motion

initial vertical KE = final PE
1/2 m vyi^2 = mg drise
vyi = sqrt (2 g drise) = sqrt (2 g (peak height - start height))

To get the jump angle
takeoff elevation = arctangent (vyi / vx)
v0y = sqrt(2g(ymax-y0)) = 3.910 m/s
t1 = sqrt(2(ymax-y0)/g) = 0.399 s
t2 = sqrt(2(ymax-y1)/g) = 0.440 s
t = t1+t2 = 0.839 s
v0x = x/t = 3.098 m/s
v0 = sqrt(v0x^2+v0y^2) = 4.988 m/s
theta = arctan(v0y/v0x) = 51.61 deg

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