SOLUTION: A punter kicks a football. Its height, h, in metres, t seconds after the kick is given by the equation h(t)=-4.9t2+18.24t+0.8 . The height of an approaching blocker’s hands is mo
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-> SOLUTION: A punter kicks a football. Its height, h, in metres, t seconds after the kick is given by the equation h(t)=-4.9t2+18.24t+0.8 . The height of an approaching blocker’s hands is mo
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Question 1156797: A punter kicks a football. Its height, h, in metres, t seconds after the kick is given by the equation h(t)=-4.9t2+18.24t+0.8 . The height of an approaching blocker’s hands is modelled by the equation b(t) = -1.43t +4.26 using the same time. Can the blocker knock down the punt? If so, at what point will it happen? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! See if the two are equal.
-4.9t^2+18.24t+0.8=-1.43t +4.26
-4.9t^2+19.67t-3.46=0
or 4.9t^2-19.67t+3.46=0
t=(1/9.8)(19.67+/- sqrt (19.67^2-4(3.46*4.9) the sqrt term=17.87 negative root only since this is the least time elapsed, and that is when punts are blocked.
t=(1/9.8)(1.8)=0.18 sec Yes
