Use the exact values you enter in previous answer(s) to make later calculation(s

Question

Use the exact values you enter in previous answer(s) to make later calculation(s).

A basketball star covers 2.60 m horizontally in a jump to dunk the ball (see figure below left). 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 1.06 m when he leaves the floor. It reaches a maximum height of 1.80 m above the floor and is at elevation 0.870 m when he touches down again. (Take the upward and forward directions to be positive.)

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

(b) Determine his horizontal velocity at the instant of takeoff. (Indicate the direction with the sign of your answer.)
m/s

(c) Determine his vertical velocity at the instant of takeoff. (Indicate the direction with the sign of your answer.)
m/s

(d) Determine his takeoff angle.

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)

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