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A punter kicks a football. Its height (h) in meters, t seconds after the kick is givenby the equation: h(t) = -4.912t^2 +18.24t +0.8. The height of an approaching blocker'shands is modeled by the equation: g(t) = -1.43t+4.26, using the same time. Can theblocker knock down the punt? If so, at what time does this happen?

1 Answer

7 votes

t=\frac{-b\pm\sqrt[]{b^2-4ac}\text{ }}{2a}
t=\frac{19.67\text{ \pm}\sqrt[]{(19.67)^2-4\cdot4.9\cdot3.46}}{2\cdot4.9}
t=(19.67\pm17.86)/(9.8)
-
t_1=3.82
t_2=\text{ 0.1}8

A punter kicks a football. Its height (h) in meters, t seconds after the kick is givenby-example-1
A punter kicks a football. Its height (h) in meters, t seconds after the kick is givenby-example-2
A punter kicks a football. Its height (h) in meters, t seconds after the kick is givenby-example-3
A punter kicks a football. Its height (h) in meters, t seconds after the kick is givenby-example-4
User Nabegh
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