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+12at2

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

After 1.0 s the balloon has not hit the ground — it will hit later. I used x = x0 + v0t + 1/2 a t^2 with v0 = 0 and a = 9.8 m/s^2 to find the distance fallen in 1.0 s: s = 1/2·9.8·(1.0)^2 = 4.9 m. Because 4.9 m < 10 m (the window height), the balloon is still 10 − 4.9 = 5.1 m above the ground; solving 10 = 1/2·9.8·t^2 gives the impact time t ≈ 1.43 s.