A basketball star covers 2.75 m horizontally in a jump to dunk the ball. His mot

Question

A basketball star covers 2.75 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 1.75 m above the floor and is at elevation 0.950 m when he touches down again.

(a) Determine his time of flight (his "hang time").
s

(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.

Explanation / Answer

First you need to figure out the flight time.

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)

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