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WILL MARK BRAINLIEST
The height (in feet) of a rocket launched from the ground is given by the function f(t) = -16t2 + 160t. Match each value of time elapsed (in seconds) after the rocket’s launch to the rocket's corresponding instantaneous velocity (in feet/second).

WILL MARK BRAINLIEST The height in feet of a rocket launched from the ground is given by the function ft 16t2 160t Match each value of time elapsed in seconds a class=

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I assume you meant h(t) = 160t - 16t^2, as h(t) = 160t - 16t2 has no maximum.

There's two ways of solving this:

(i) By completing the square and finding the maximum turning point.

(ii) By using Calculus methods to find derivative and equating it to 0 in order to find maximum turning point.

(i) h(t) = 160t - 16t^2

h(t) = -16t^2 +160t

h(t) = -16(t^2-10t) (by taking out a common factor of -16)

h(t) = -16[(t-5)^2 - 25] (by completing the square)

h(t) = -16(t-5)^2 + 400 (on multiplying out by -16)

From this we see that the turning point is at (5;400), therefore the maximum height is 400 feet and is reached after 5 seconds.

Or...

(ii) h(t) = 160t - 16t^2

h`(t) = 160 - 32t (where h`(t) = derivative of h(t))

Now to find maximum of h(t), we set h`(t) = 0 and solve for t:

0 = 160 - 32t (on substituting h`(t) = 0)

32t = 160 (on solving for t)

t = 160/32 (on dividing both sides by 5)

t= 5

Now we have found that at 5 seconds, we will reach our maximum height. So to find this maximum height, we'll have to substitute t=5 into h(t) = 160t - 16t^2.

h(5) = 160(5) -16(5)^2

h(5) = 800 - 400

h(5) = 400 feet

So, once again we have shown that maximum height is 400 feet and is reached after 5 seconds.

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