A rock is thrown upward from a bridge that is 87 feet above

A rock is thrown upward from a bridge that is 87 feet above a road. The rock reaches its maximun above the road 0.85 seconds after it is thrown and contacts the road 2.99 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. height

Solution

constant term of quadratic equation is 87 as that is the height above the road

x coordinate of vertex is 0.85

standard quadratic equation is given by

H(t) = at^2 + bt + c

c = 87

-b/2a = 0.85

b = -1.7a

0 = a(2.99)^2 + b(2.99) + c

0 = a(2.99)^2 + b(2.99) + 87

0 = a(2.99)^2 -1.7a (2.99) + 87

0 = 8.9401a - 5.083a + 87

a = -22.556

b = 38.345

hence, quadratic equation is

f = -22.556t^2 + 38.245 t + 87