How do I solve this...simple quadratic equation? 10pts?

t = -64/((2)(-16)) = 2 s.

h(2) = 23 + 128 - 64 = 87 ft.

h = 23 + 64t - 16t²
h = 23 - 16(t² - 4t + 2²) + 16(2²)
h = 23 - 16(t - 2)² + 64
h = 87 - 16(t - 2)²

For any real number t, - 16(t - 2)² ≤ 0
Thus, h = 87 - 16(t - 2)² ≤ 87

h is the maximum when at t = 2
max. of h = 87

The arrow reaches the maximum in 2 seconds. ...... (Ans)
The maximum height is 87 meters. ...... (Ans)

h = 23 + 64t - 16t²
dh/dt = 64 - 32t = 32(2 - t)
d²h/dt² = -32 < 0

When t = 2 :
h = 23 + 64(2) - 16(2)² = 87
dh/dt = 0
d²h/dt² < 0

The maximum height is 87 meters in 2 seconds.

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.12 s later. You may ignore air resistance.

A) let the initial speed v0 of the first ball be given and treat the height h of the building as an unknown. What must the height of the building be for both balls to reach the ground at the same time for v0 = 8.75 m/s .

B) If v0 is greater than some value vmax, a value of h does not exist that allows both balls to hit the ground at the same time. Solve for v -max.

C) If v0 is less than some value vmin, a value of h does not exist that allows both balls to hit the ground at the same time. Solve for v-min.