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:

Question

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. 

(4 points)

Answer 1

To determine whether the balloon would hit the ground before or after 1.0 seconds of falling, I used the first equation of motion: \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \). In this case, the initial position \( x_0 \) is 10 meters, the initial velocity \( v_0 \) is 0 m/s (since the balloon is dropped), and the acceleration \( a \) is -9.8 m/s² (it's falling downward). Plugging in these values for \( t = 1.0 \) seconds, we get:

\[ x = 10 + 0 \cdot 1 + \frac{1}{2} \cdot (-9.8) \cdot (1)^2 = 10 - 4.9 = 5.1 \text{ meters}. \]

Since the position after 1 second (5.1 meters) is still above the ground (0 meters), the balloon has not yet hit the ground. Therefore, it will take longer than 1.0 seconds to reach the ground.

Answer 2

What’s the answer

Answer 3

The balloon will not hit the ground before 1.0 seconds of falling. I used the equation \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \) to calculate the position of the balloon after 1 second. Substituting the values, I found that after 1 second, the balloon would be at a height of 5.1 meters, which is still above the ground (0 meters). Thus, it will take longer than 1.0 seconds for the balloon to hit the ground.