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.
To determine whether the balloon would hit the ground before or after 1.0 second of falling, we can use the equation x = x0 + v0t + (1/2)at^2, where x is the final position, x0 is the initial position, v0 is the initial velocity, t is time, a is acceleration, and ^2 represents squared.
Given that x0 = 10 meters, a = 9.8 m/s^2, and we want to find the value of t when x = 0 (since the balloon would have hit the ground), we can plug these values into the equation.
0 = 10 + 0t + (1/2)(9.8)t^2
0 = 10 + (4.9)t^2
Simplifying the equation, we get:
t^2 = -2.0408
Since time cannot be negative, we can conclude that the balloon would not hit the ground before 1.0 second of falling. Therefore, the balloon will hit the ground after 1.0 second.
Based on this reasoning, the equation x = x0 + v0t + (1/2)at^2 was used to determine that the balloon would not hit the ground by then.