A rock is thrown upward from a bridge that is 88 feet above a road. The rock reaches its maximum height above the road 0.67 seconds after it is thrown and contacts the road 3.18 seconds after it was thrown. Define a quadratic function, f, that gives the height of the rock above the road (in feet) in terms of the number of seconds elapsed since the rock was thrown, t

Answer 1

you know that if v is the initial upward velocity, v-9.8*0.67 = 0

Plug that into the usual equation

h(t) = 88 + 6.566t - 4.9t^2

Then check that h(3.18) = 0

Or, since a parabola is symmetric, you know that

h(t) = a(t-0.67)^2 + 88
h(3.18) = 0, so
a(3.18-0.67)^2 + 88 = 0
h(t) = -13.97(t-0.67)^2+88

Hmmm. The two functions don't match, but the second fits. Maybe you can check my math.