An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h (t) = - 16t2 + 64t + 80?

1) we calculate the first derivative:

h' (x) = - 32t+64

2) we equalized to "0" the first derivative, and find out the value of "t".

-32t+64=0

-32t=-64

t=-64/-32

t=2

3) we calculate the second derivative:

h''=-32<0 ⇒ then, we have a maximum at t = (2)

4) we calculate the height at t=2

h (2) = - 16 (2) ²+64 (2) + 80=-32+128+80=176.

Answer: the maximum height is 176 m.

Ht + 16t2 - 64t - 80 = 0

ht - (((0 - 24t2) + 64t) + 80) = 0

3.1 Solve ht+16t2-64t-80 = 0

theres no solution found honey