92_Physics ProblemsTechnical Physics

92_Physics ProblemsTechnical Physics - 93 Chapter 4 P4.24...

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Chapter 4 93 P4.24 From the instant he leaves the floor until just before he lands, the basketball star is a projectile. His vertical velocity and vertical displacement are related by the equation vv a y y yf yi y f i 22 2 =+ di . Applying this to the upward part of his flight gives 02 9 8 0 1 8 5 1 0 2 2 v yi .. . ms m 2 ej af . From this, v yi = 403 . m s. [Note that this is the answer to part (c) of this problem.] For the downward part of the flight, the equation gives v yf 2 0 2 9 80 0 900 1 85 =+ − . m 2 a f . Thus the vertical velocity just before he lands is v yf =− 432 . m s . (a) His hang time may then be found from vva t yf yi y : −= + 980 . 2 t or t = 0852 . s . (b) Looking at the total horizontal displacement during the leap, xv t xi = becomes 280 m s = v xi which yields v xi = 329 m s . (c) v yi = 4.03 m s . See above for proof.
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