Need Help? ReadItWatch 11.-/15 points LarPCalc10 2.1068 The path of a punted foo

Question

Need Help? ReadItWatch 11.-/15 points LarPCalc10 2.1068 The path of a punted football is modeled by 16 1225 where f(x) is the height (in feet) and x is the horizontal distance (in feet) from the point at which the bal is punted. a) How high is the ball when it is punted? (b) What is the maximum height of the punt? (Round your answer to two decimal places.) ft (e) How long is the punt? (Round your answer to two decimal places.) ft Need Help? SaveProgress Practice Another Version Submit Answer: t O Type here to search

Explanation / Answer

f(x)=-(16/1225)x2+(7/5)x+1.5

(a)

initially x=0

=>f(0)=-(16/1225)02+(7/5)0+1.5

=>f(0) =1.5

ball is 1.5ft high when it is punted

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(b)

f(x)=-(16/1225)x2+(7/5)x+1.5

f(x)=-(16/1225)(x2-(7/5)(1225/16)x) +1.5

f(x)=-(16/1225)(x2-(1715/16)x) +1.5

f(x)=-(16/1225)(x2-(1715/16)x+(-1715/32)2-(-1715/32)2) +1.5

f(x)=-(16/1225)(x2-(1715/16)x+(-1715/32)2)+((16/1225)(-1715/32)2) +1.5

f(x)=-(16/1225)(x-(1715/32))2 +(2401/64)+1.5

f(x)=-(16/1225)(x-(1715/32))2 +(2497/64)

maximum height =(2497/64) ft

maximum height =39.02  ft

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c)

at the end ball reaches the ground. then f(x)=0

=>-(16/1225)(x-(1715/32))2 +(2497/64)=0

=>(16/1225)(x-(1715/32))2 =(2497/64)

=>(x-(1715/32))2 =(2497/64)(1225/16)

=>(x-(1715/32))2 =(3058825/1024)

=>(x-(1715/32))=(3058825/1024)

=>x=(1715/32)+(3058825/1024)

=>x=108.25 ft

punt is 108.25 ft long

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