# 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/s^2. There are two equations that cam be used to describe its motion over time:

1. x = x0 + v0t + (1/2)at^2

2. 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.

(I'm pretty sure it's the first equation, but I don't know what numbers to plug in where. I also don't understand the "what comparison did you make" part.)

## To determine whether the balloon hits the ground before or after 1.0 s of falling, we can use the first equation, x = x0 + v0t + (1/2)at^2. In this equation, x represents the final position (in this case, the ground), x0 is the initial position (10 meters above the ground), v0 is the initial velocity (0 m/s as the balloon was initially at rest), a is the acceleration (-9.8 m/s^2), and t is the time. By plugging in these values into the equation, we can solve for t. If the calculated t is less than 1.0 s, it means that the balloon hits the ground before 1.0 s. If the calculated t is greater than 1.0 s, it means the balloon hits the ground after 1.0 s. The comparison made is whether t is less than or greater than 1.0 s.

## To determine whether the balloon hits the ground before or after 1.0 s of falling, we can use Equation 1: x = x0 + v0t + (1/2)at^2.

Let's assign the following values:

- x0 (initial position) = 10 meters (height of the window)

- v0 (initial velocity) = 0 m/s (the balloon is initially at rest)

- a (acceleration due to gravity) = 9.8 m/s^2 (given)

We want to find the time, t when the balloon hits the ground (x = 0 meters). In this case, x0 = 10 meters and x = 0 meters. Plugging these values into the equation, we have:

0 = 10 + 0t + (1/2)(9.8)t^2

Simplifying this equation, we get:

4.9t^2 = 10

To solve for t, divide both sides of the equation by 4.9:

t^2 = 10/4.9

Taking the square root of both sides, we find:

t ≈ 1.02 seconds

From this calculation, we can see that the balloon hits the ground just after 1.0 second. Therefore, using Equation 1, we determined that the balloon hits the ground after 1.0 s of falling.

The "comparison" part of the question refers to comparing the time calculated (1.02 seconds) with the given time (1.0 second). By comparing the calculated time to the given time, we can determine whether the balloon hits the ground before or after 1.0 s of falling.

## yes ... 1st equation

plug in 1.0 s for t , and see how far the balloon has fallen

x is the distance fallen

x0 and v0 are zero ... the balloon starts from rest

a is gravitational acceleration

the comparison is between the starting height

... and the distance fallen in 1.0 s