At the circus, a poodle is shot out of a cannon. The height above the ground (in feet) of the dog after t seconds is given by `f(t)=-16t^2+78t+6` . Round answers to the nearest hundredth. a. When...

At the circus, a poodle is shot out of a cannon. The height above the ground (in feet) of the dog after t seconds is given by `f(t)=-16t^2+78t+6` . Round answers to the nearest hundredth.


a. When will the poodle land in a pool of water at ground level?


b. Give the time interval that the poodle’s height is more than 20 ft.


c. What is the poodle’s maximum height during flight?

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tjbrewer | Elementary School Teacher | (Level 2) Associate Educator

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a) `-16t^2+78t+6=0` we find t value by applying the quadratic formula `t=(-78+-sqrt(78^2-4(-16)(6)))/((2)(-16))=(-78+-sqrt(6084+384))/(-32)=`

`(-78+-sqrt(6468))/(-32)=(-78+-80.42)/(-32)=2.42/-32, -158.42/-32=-.08, 4.95` Common sense tells us that t cannot have a negative value, so t=4.95 sec. 

b)` ``-16t^2+78t+6>20` We solve this by subtracting 20 from both sides, and then re-applying the quadratic formula to the inequality `-16t^2+78t-14>0` The interval is between the two values of t for `t=(-78+-sqrt(78^2-4(-16)(-14)))/((2)(-16))=(-78+-sqrt(6084-896))/(-32)=`

`(-78+-sqrt(5188))/-32=(-78+-72.03)(-32)=(-5.97)/-32, (-150.03)/-32=`

`0.19, 4.69`So the poodle's height exceeds 20 ft from 0.19sec, to 4.69 sec. 

c) The formula for the Max/min of a quadratic equation is `t=-(78/(-32))=2.44` The poodle will reach max height at 2.44sec, and plugging that value of t into our equation finds a maximum height.  `f(2.44)=-16(2.44)^2+78(2.44)+6=`

`-95.26+190.32+6=` 101.06 ft. 

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