A rock is thrown upward from a bridge that is 45 feet above a road. The rock rea

Question

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.

Explanation / Answer

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

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