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

v=v0+at

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.

Answer 1

To determine if the balloon will hit the ground before or after 1.0 second of falling, we can use the equation x = x0 + v0t + 1/2at^2, where x is the final position, x0 is the initial position (10 meters), v0 is the initial velocity (0 m/s), and a is the acceleration due to gravity (-9.8 m/s^2).

We substitute the values into the equation:
x = 10 + (0)(1) + 1/2(-9.8)(1)^2
x = 10 - 4.9
x = 5.1 meters

Since the final position after 1.0 second is 5.1 meters, which is less than the height of the window (10 meters), the balloon would not hit the ground by then. Therefore, the balloon will hit the ground after 1.0 second of falling.