Question

A student holds a water balloon outside of an open window and lets go. The window is 10 meters above the ground, and the balloon is falling under the acceleration of gravity, which is 9.8 m/s2. There are two equations that can be used to describe its motion over time
x=x0+v0t+1/2at^2

Would the balloon hit the ground before or after 1.0 s of falling? Which equation did you use to decide, and what comparison did you make to determine that it would or would not hit the ground by then?

Use 3–5 sentences to explain your reasoning.

(4 points)

Answer 1

Use the kinematic equation x = x0 + v0 t + 1/2 a t^2 with v0 = 0. After t = 1.0 s the fall distance is (1/2)(9.8)(1.0)^2 = 4.9 m. Since 4.9 m < 10 m, the balloon has not yet reached the ground at 1.0 s. (For reference, the impact time is t = sqrt(2·10/9.8) ≈ 1.43 s, which is after 1.0 s.)