Homework SourseA rock is thrown upward from a bridge that is 45 feet above
A rock is thrown upward from a bridge that is 45 feet above a road. The rock reaches its maximum height above the road 0.83 seconds after it is thrown and contacts the road 2.53 seconds after it was thrown. Define a quadratic function, f, that gives the height of the rock above the ground (in feet) in terms of the number of seconds elapsed since the rock was thrown, t.
A rock is thrown upward from a bridge that is 45 feet above a road. The rock reaches its maximum height above the road 0.83 seconds after it is thrown and contacts the road 2.53 seconds after it was thrown. Define a quadratic function, f, that gives the height of the rock above the ground (in feet) in terms of the number of seconds elapsed since the rock was thrown, t.
Solution
Solution:
You know that if v is the initial upward velocity,
v-9.8*0.83 = 0
Substitute that into the usual equation
h(t) = 45 + 8.134t - 4.9t2
Then check that h(2.53) = 0
Or, since a parabola is symmetric, you know that
h(t) = a(t-0.83)2 + 45
h(2.53) = 0,
So
a(2.53-0.83)2 + 45 = 0
h(t) = -13.97(t-0.83)2 + 45
