Question

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?

Answer 1

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.

Answer 2

ht + 16t2 - 64t - 80 = 0
ht - (((0 - 24t2) + 64t) + 80) = 0
3.1 Solve ht+16t2-64t-80 = 0
theres no solution found honey