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/s^2. 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?
In 5 sentences explain your reasoning and explain as if you were an 11th grader.
As an 11th grader, I can use the second equation, v = v0 + at, to determine if the balloon hits the ground before or after 1.0 s of falling. I know that the initial velocity, v0, is 0 because the balloon is just released and not already moving. The acceleration, a, is given as -9.8 m/s^2 because gravity pulls objects downward.
If I substitute the values into the equation, it becomes v = 0 + (-9.8)(t). Simplifying further, v = -9.8t.
Now, I can compare this equation with the condition for the balloon to hit the ground before 1.0 s. The balloon hits the ground when its position, x, becomes 0.
The first equation, x = x0 + v0t + (1/2)at^2, becomes 0 = 0 + 0 + (1/2)(-9.8)(t^2). Simplifying, 0 = -4.9t^2.
Since the equation gives us 0 for the position, it suggests that the balloon does reach the ground in time. Therefore, we can conclude that the balloon hits the ground before 1.0 s of falling.