U3.L4.HW: Quadratic Functions: Question 27

Question

How do I figure this out?:

NASA launches a rocket at t = 0 seconds. Suppose its height, in meters above sea-level, as a function of time is given by h(t) = -4.9t² + 112t + 227.

How high above sea-level does the rocket get at its peak? Round your answer to 2 decimal places.

The rocket peaks at ​__________ meters above sea-level.

Answer 1

h = -4.9t^2 + 112t + 227

solve with calculus

take the derivative and set = 0

velocity = h'(t) = v(t) = -9.8t + 112 = 0

t = 112/9.8 = about 11.43 seconds

to reach maximum height

max h = -4.9(112/9.8)^2 +112(112/9.8) + 227

= -(112)^2/2(9.8) + 2(112)/2(9.8) +227

= -640 +1280+227

= 640 + 227

= 867 meters maximum height

or

solve with algebra

h = -4.9(t^2 - 112t/4.9) + 227

= -4.9(t^2 - 112t/4.9 + (112/9.8)^2) + 227 + 4.9(112/9.8)^2

= -4.9(t - 112/9.8)^2 + 227+ 640

= -4.9(t-112/9.8)^2 + 867

in vertex form

vertex = (112/9.8, 867) = maximum point

with 867 meters = maximum height

Answer 2

Hi Manny Y.

Your function above is in the form of the Quadratic Formula

h(t) = -4.9t²+112t +227

f(x)= ax²+bx+c

You can look this up online or in your text and confirm the following

When a is negative the parabola opens downward so the vertex would represent the highest point.

Also the x=h value of the vertex is -b/2a once this is determined you plug it back into your function to find the y=k value of your vertex and the highest point of your parabola. Your highest point will be found in the coordinates of your vertex. Also you should graph your function to confirm your vertex. You can graph your function at Desmo.com

Can you continue from here?

I hope this help is you have questions just write back.