106 Motion in Two DimensionsP4.58The football travels a horizontal distanceRvgii==°=22220 060 098035 3sin.sin...θbgaf afm.Time of flight of ball istvg°=22200300204sin(. )sin...s.FIG. P4.58The receiver is ∆xaway from where the ball lands and ∆x=−=35 320 015 3... m. To cover thisdistance in 2.04 s, he travels with a velocityv15 3750...m s in the direction the ball was thrown .P4.59(a)∆ygt=−122; ∆xvti=Combine the equations eliminating t:∆∆xviFHGIKJ122.From this, ∆∆xygvi222=−FHGIKJFIG. P4.59thus ∆∆ygi=−=−−
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