Question

A student holds a water balloon outside of an open window and let's 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.
2. V= vo + at

Answer 1

The motion of the water balloon can be described by the equations of motion under constant acceleration. In this case, the acceleration is due to gravity (approximately \(9.8 \, \text{m/s}^2\) downward). Given that the balloon is released from rest, the initial velocity (\(v_0\)) is \(0 \, \text{m/s}\).

### 1. Velocity Equation

You provided the velocity equation:

\[
v = v_0 + at
\]

Substituting the initial conditions:

- \(v_0 = 0 \, \text{m/s}\) (the initial velocity)
- \(a = 9.8 \, \text{m/s}^2\)

The equation simplifies to:

\[
v = 0 + (9.8 \, \text{m/s}^2)(t) = 9.8t
\]

### 2. Displacement Equation

The displacement (\(s\)) of the balloon from the point of release can be represented by the equation:

\[
s = v_0 t + \frac{1}{2} a t^2
\]

Again substituting the initial conditions (\(v_0 = 0\)):

\[
s = 0 + \frac{1}{2}(9.8 \, \text{m/s}^2)t^2 = 4.9t^2
\]

### Total Displacement from the Window to the Ground

Given that the window is \(10 \, \text{m}\) above the ground, the total displacement equation from the point of release until it hits the ground would be:

\[
10 = 4.9t^2
\]

### Find Time (\(t\))

Rearranging the equation to solve for \(t\):

\[
t^2 = \frac{10}{4.9}
\]

Calculating this:

\[
t^2 \approx 2.04
\]
\[
t \approx \sqrt{2.04} \approx 1.43 \, \text{s}
\]

### Conclusion

The water balloon takes approximately \(1.43\) seconds to reach the ground from a height of \(10\) meters. You can also find the final velocity just before it hits the ground by substituting \(t\) back into the velocity equation:

\[
v = 9.8t = 9.8(1.43) \approx 14.0 \, \text{m/s}
\]

Therefore, the water balloon hits the ground at a speed of approximately \(14.0 \, \text{m/s}\).