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 will hit the ground before or after 1.0 second of falling, let's consider the second equation v = v0 + at. The initial velocity (v0) of the balloon is 0 m/s since it was just released. The acceleration (a) is the acceleration due to gravity, which is 9.8 m/s^2, and the time (t) is 1.0 second.
Plugging in these values, we get v = 0 + 9.8 * 1.0, which simplifies to v = 9.8 m/s. This means that after 1.0 second of falling, the balloon will have a velocity of 9.8 m/s.
Now, if we use the first equation x = x0 + v0t + 0.5at^2, we can calculate the displacement (x) of the balloon after 1.0 second. The initial position (x0) is 10 meters above the ground, the initial velocity (v0) is 0 m/s, the acceleration (a) is -9.8 m/s^2 (taking downward as negative), and the time (t) is 1.0 second.
Substituting these values, we have x = 10 + 0 + 0.5 * (-9.8) * (1.0)^2. Simplifying this equation gives us x = 10 - 4.9, which results in x = 5.1 meters.
Since the displacement is equal to 5.1 meters after 1.0 second, and the balloon started at a height of 10 meters, we can conclude that the balloon has not hit the ground yet after 1.0 second. Therefore, it will hit the ground after more than 1.0 second of falling.