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+12at2x=x0+v0t+12at2 v=v0+atv=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 if the balloon will hit the ground before or after 1.0 s of falling, we can use the equation x=x0+v0t+1/2at^2. We know that x0 (initial position) is 10 m and a (acceleration) is -9.8 m/s^2 (due to gravity). Since the balloon is falling downwards, the initial velocity v0 would be 0 m/s. Plugging these values into the equation, we can calculate the position of the balloon after 1.0 s.
x = 10 + 0(1.0) + 1/2(-9.8)(1.0)^2
x = 10 - 4.9
The result is -4.9 m, which means the balloon would hit the ground after 1.0 s of falling.