Chapter 493P4.24From the instant he leaves the floor until just before he lands, the basketball star is a projectile. Hisvertical velocity and vertical displacement are related by the equation vvayyyfyiyfi222=+−di.Applying this to the upward part of his flight gives 02980 1851022−−vyi...msm2ejaf. From this,vyi=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 vyf202 9 800 900 1 85=+ −−.m2af.Thus the vertical velocity just before he lands isvyf=−432. ms.(a)His hang time may then be found from vvatyfyiy:−=+−980.2tor t=0852. s.(b)Looking at the total horizontal displacement during the leap, xvtxi=becomes280m s=vxiwhich yields vxi=329ms.(c)vyi=4.03 m s . See above for proof.
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