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Jake tosses a coin up in the air and lets it fall to the ground. The equation that models the height (in feet) and time (in seconds) of the parabola is h(t)= -16t^2+ 24t+6. What is the height of the coin when Jake tosses it?

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Answer:

The maximum height of coin Jake tosses is 15 feet.

Explanation:

Given Jake tosses a coin up in the air

and the height of coin is model by h(t)=
(-16)t^(2) +24t+6

To find height of the coin when Jake tosses it:

When a coin is in the hand of Jake, time t=0

Height of coin is h(t)=
(-16)t^(2) +24t+6

h(0)=
(-16)(0)^(2) +24(0)+6

h(0)=6 feet.

Therefore, Height of coin at time t=0 is 6.

For maximum height of the coin,

h(t)=
(-16)t^(2) +24t+6

Differentiating both side,


(d)/(dt)h(t)=(d)/(dt)[(-16)t^(2) +24t+6]


(d)/(dt)h(t)=(d)/(dt)[(-32)t+24]


(d)/(dt)h(t)=0


(d)/(dt)[(-32)t+24]=0

t=
(24)/(32)

t=
(3)/(4)

t=0.75

Now,

h(t)=
(-16)t^(2) +24t+6

h(0.75)=
(-16)(0.75)^(2) +24(0.75)+6

h(0.75)=15 feet.

The maximum height of coin jake tosses is 15 feet.

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