Asmall fish is dropped by a pelican that is rising steadily at 0.50 m/s. how far below the pelican is the fish after 2.5 s? provide the formula.
a) 61 m
b) 29.3 m
c) 30.6 m
d) 1.1 m

The initial velocity of the fish is the same as that of the pelican and it is -0.50 m/s. The distance traveled by the object is given by the equation,
                                 d = (V1)t + 0.5at²
where t is equal to 2.50s and the acceleration due to gravity, a, is 9.80 m/s². Substituting,
                                 d = (-0.50 m/s)(2.5 s) + 0.5(9.8 m/s²)(2.50 s)²
                                     d = 30.625 m
Thus, the answer is letter C.