Given height function
We have to find the maximum height of the projectile.
To get the critical point, we put
So, the critical point is
So, by second derivativetest, h(t) has maximum value at
Henca, the maximum height of the projective is 256 feet.
complete the square
This is an equation of a parabola that opens upward with vertex at (2,64) maximum height of the projectile
For this case we have the following function:
Deriving the function we have:
h'(t)=-32t + 128
Equaling zero we have:
-32t + 128 = 0
We clear the time:
t = 4
Substituting in the height equation:
h(4) = 256 feet
The maximum height is:
h (4) = 256 feet
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