Final answer:
The parametric equations for the water balloons dropped from a 200 feet tall building are: X1 = 0, X2 = 0, Y1(t) = -16t^2 + 200, and Y2(t) = -16t^2 - 40t + 200. These equations account for the initial height of the building, the acceleration due to gravity, and the initial velocity of the second balloon.
Step-by-step explanation:
The student is asked to write parametric equations to represent the path of two water balloons, one dropped and one thrown from a building. Let's denote the first water balloon that is simply dropped as balloon 1 and the second water balloon that is thrown straight down as balloon 2.
Since the horizontal motion is not involved here, the horizontal position for both balloons remains constant over time. Therefore, for both water balloons, X1 and X2 will just be equal to the horizontal position from which they were dropped (or thrown), which we'll assume to be at the origin, so X1 = 0 and X2 = 0.
The vertical positions Y1 and Y2 are influenced by the initial velocity and the acceleration due to gravity. For balloon 1, which is dropped with no initial velocity, and assuming gravity g=32 ft/s2, the parametric equation for its vertical motion is given by Y1(t) = -16t2 + 200, where 200 feet is the initial height of the building. For balloon 2, which is thrown downward with an initial velocity of 40 ft/s, the parametric equation for its vertical motion is Y2(t) = -16t2 - 40t + 200.