A basketball star covers 2.70 m horizontally in a jump to dunk the ball. His mot

Question

A basketball star covers 2.70 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 1.80 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")._____ 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._____° 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 = 2.40 m, and yf = 0.800 m. _____ s

Explanation / Answer

a) Using v² vo² = 2g h we can find the velocity

For the first part of the jump v= (2 ×9.8 ×0.78) =3.909 m/s. This actually is the vertical component of initial (take-off) velocity. For the touch-down velocity, v = (2 × 9.8 × 0.89) = 4.176 m/s.

Now lets find time using v vo = gt.Rise time = 3.909/9.8 = 0.3988 sec.Fall time = 4.176/9.8 = 0.4261 sec.

Thus, Hanging time= rise time + fall time = 0.8249 sec.

b) The horizontal component of initial velocity is found from x = vt. Horizontal motion is not accelerated, so velocity remains constant. v = 2.7/0.8249 = 3.273 m/s.

c) This was found above: 3.909 m/s, for take-off.

d) = atan 3.909/3.273 = 50.06°.

e) Rise time = 0.4947 sec. Fall time = 0.5714 sec

Hang time=1.066 sec

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