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.

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.
1. X= x0+ vot+ 1/2 at 2.

GPT-4o mini · Answer

To analyze the motion of the water balloon falling from a height of 10 meters, we can apply the kinematic equation you provided:

$X = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$

where:

$X$ is the final position
$x_{0}$ is the initial position (height from which the balloon is dropped)
$v_{0}$ is the initial velocity of the balloon
$t$ is the time in seconds
$a$ is the acceleration due to gravity (approximately -9.8 m/s²; negative because it acts downwards)

In this scenario:

The initial position $x_{0} = 10$ meters (the height of the window)
The initial velocity $v_{0} = 0$ m/s (the balloon is released and not thrown)
The acceleration $a = - 9.8$ m/s² (downward)

Given these values, we can rewrite the equation as:

$X = 10 + 0 \cdot t + \frac{1}{2} (- 9.8) t^{2}$

This simplifies to:

$X = 10 - 4.9 t^{2}$

Time to Reach the Ground

To find out when the balloon hits the ground $X = 0$ :

$0 = 10 - 4.9 t^{2}$

Rearranging gives:

$4.9 t^{2} = 10$

Now solving for $t^{2}$ :

$t^{2} = \frac{10}{4.9} \approx 2.04$

Taking the square root:

$t \approx \sqrt{2.04} \approx 1.43 seconds$

Summary

The balloon will take approximately 1.43 seconds to hit the ground after being released from the window.

You can use the modified equation to find the position $X$ of the balloon at any given time $t$ as it falls. Just plug in your desired value for $t$ .