A toy rocket is launched from the top of 48-foot hill. The rocket's initial upward velocity if 32 feet per second and its height h at any given second t is modeled by the equation h=- 16t^2+32t+48.

A: To find out how long the rocket was in the air, we need to calculate when the height h reaches 0 (ground level). So we set h=0 in the equation and solve for t:

-16t^2 + 32t + 48 = 0
Divide by -16 to simplify:
t^2 - 2t - 3 = 0
Now we need to solve this quadratic equation for t. We can factor it:
(t-3)(t+1) = 0
This gives us two possible values for t: t=3 and t=-1. Since time cannot be negative, we discard t=-1.

Therefore, the rocket was in the air for 3 seconds.

B: To find how high the rocket was at 2 seconds, we simply substitute t=2 into the equation for h:
h = -16(2)^2 + 32(2) + 48
h = -64 + 64 + 48
h = 48 feet

So, the rocket was 48 feet high at 2 seconds.

C: To find out how high the rocket got, we need to determine its maximum height. The maximum height occurs at the vertex of the parabolic function, and the formula for finding the vertex of a parabola is given by: t = -b / 2a.

In our case, a = -16 and b = 32. Substituting these into the formula:
t = -32 / 2(-16)
t = -32 / -32
t = 1

So, the rocket reached its maximum height at 1 second. To find out how high it got, we simply substitute t=1 into the equation for h:
h = -16(1)^2 + 32(1) + 48
h = -16 + 32 + 48
h = 64 feet

Therefore, the rocket reached a maximum height of 64 feet.