Need Help ReadItWatch 1115 points LarPCalc10 21068 The path
Solution
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

