Question

A flare was launched straight up from the ground with an initial velocity of 176 ft/s and returned to the ground after 11 s.

The height of the flare t seconds after launch is modeled by the function f(t)=−16t2+176t .

What is the maximum height of the flare, in feet?

Answer 1

One way to figure out the maximum height is to take the derivative of the equation.
So, f'(t)=-32t+176
Find for what values of t f'(t) is 0, as that is the maximum/minimum of your function (when the slope of the tangent is 0).
So 0=-32t+176
-32t=-176
t=5.5 s.
So at 5.5 s is when a max height takes place. Plug it back into your original equation, and you should get:
f(5.5)=-16(5.5)^2+176(5.5)
f(5.5)=484 ft

Another way to get it is by doing this:
The maximum height must occur at it's vertex, which should be in the middle of your x-intercepts.
Knowing that f(t)=-16t^2+176t, you can factor.
0=t(-16t+176)
So the first x intercept is at 0, and the second is at x=11.
To get the middle point, add up the x intercepts and divide it by 2, and you should get that x=5.5. Plug that back into your equation and you should have the maximum height.

Answer 2

@pandasurvive

Answer 3

Just wondering if it was 16x^2 wouldnt it be 16x time 16x instead of 32x?

Answer 4

@callielovesyhuu

Answer 5

f'(t)=-32t+176
Find for what values of t f'(t) is 0, as that is the maximum/minimum of your function (when the slope of the tangent is 0).
So 0=-32t+176
-32t=-176
t=5.5 s.
So at 5.5 s is when a max height takes place. Plug it back into your original equation, and you should get:
f(5.5)=-16(5.5)^2+176(5.5)
f(5.5)=484 ft

I swear, teachers use this one for homework all the freaking time.

Answer 6

@psirockin2 just wondering are you sure?