Asked by David
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
Answers
Answered by
Steve
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.
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.
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