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

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?

Answer 1

To determine whether the water balloon would hit the ground before or after 1.0 second of falling, we can use the kinematic equation for an object in free fall under the influence of gravity. One suitable equation is:

$$h(t)=h_0-\frac{1}{2}g{t}^2$$

where:

• $h(t)$ is the height of the object at time $t$,
• $h_0$ is the initial height (10 meters in this case),
• $g$ is the acceleration due to gravity (approximately 9.8 m/s²), and
• $t$ is the time in seconds.

We want to find out if the balloon will hit the ground (0 meters) before 1.0 s. We can calculate the height of the balloon after 1.0 s:

1. Set $h_0=10$ m and $t=1.0$ s in the equation:

$$h(1.0)=10-\frac{1}{2}\cdot 9.8\cdot (1.0{)}^2$$

2. Calculate the value:

$$h(1.0)=10-\frac{1}{2}\cdot 9.8\cdot 1=10-4.9=5.1\text{ meters}$$

Since $h(1.0)=5.1$ meters, the water balloon has not yet hit the ground after 1.0 second of falling; it is still 5.1 meters above the ground.

Thus, the water balloon would hit the ground after 1.0 second of falling.