Question

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.

Answer 1

To determine whether the water balloon would hit the ground before or after 1.0 seconds, I used the first equation of motion: \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \). Given that the initial position (\( x_0 \)) is 10 meters (the height of the window), the initial velocity (\( v_0 \)) is 0 m/s (since the balloon is dropped), and the acceleration (\( a \)) is -9.8 m/s² (acting downward), we can substitute these values into the equation.

Calculating for \( t = 1.0 \) seconds:
\[ x = 10 + 0 \cdot 1 + \frac{1}{2}(-9.8)(1.0^2) \]
\[ x = 10 - 4.9 = 5.1 \text{ meters} \]

Since the position \( x \) is still above ground level (5.1 meters) after 1 second, I conclude that the balloon will not hit the ground until after this time.