A rock is thrown upward from a bridge that is 43 feet above a road. The rock reaches its maximum height above the road 0.91 seconds after it is thrown and contacts the road 3.06 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

43 + (8.93) * t + (9.81) t²/2 for 0 < t < 0.91

55 - (9.81) t²/2 for 0.91 < t < 3.06 sec

Step-by-step explanation:

Simple kinematic equations used:

a = dv/dt ... Eq 1 and v = ds/dt ... Eq 2

Using Eq 1:

a. dt = dv - separating variables

∫a. dt = ∫dv - integrating both sides 0< t < t and vi < v < vf

(a. t) = (vf - vi) - Solving and evaluating integral

Hence,

vf = vi + a*t ... Eq 3

Using Eq 2:

vf. dt = ds - separating variables

∫ (vi+a*t). dt = ∫ds - Substitute Eq 3 and integrating both sides 0< t < t and si < s < sf

vi*t + at²/2 = sf-si ... Eq 4

Now using the data given and Eq 4 & Eq 3:

Interval 1: For rock at max height above road @ t = 0.91 sec and vf = 0, for 0 < t < 0.91

vi = - a*t = - (-9.81) * (0.91) = 8.93 m/s

Plugging in Eq 4:

sf = si + vi*t + at²/2 = 43 + (8.93) * t + (9.81) t²/2 = 55 fts @ t = 0.91 sec

Interval 2: For rock in contact with road @ t = 3.06 sec and vi = 0, for 0.91 < t < 3.06

Using Eq 4

sf = si + vi*t + at²/2 = 55 - (9.81) t²/2 for 0.91 < t < 3.06 sec