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.
(4 points)
To determine whether the balloon hits the ground before or after 1.0 s of falling, we can use the equation x=x0+v0t+12at2. By plugging in the values, we get x=10+0+12(9.8)(1)^2=10+4.9=14.9. This equation gives us the position of the balloon after 1 second of falling. Since the window is at a height of 10 meters, and the balloon would be at a height of 14.9 meters after 1 second, we can conclude that the balloon would hit the ground after 1 second of falling.
To determine whether the balloon hits the ground before or after 1.0 s of falling, we can use the equation v = v0 + at. In this equation, v is the final velocity, v0 is the initial velocity, a is the acceleration due to gravity (which is -9.8 m/s^2 since it is acting downwards), and t is the time in seconds. We can rearrange this equation to solve for t:
t = (v - v0) / a
Since the initial velocity v0 is zero, the equation simplifies to t = v / a. The initial velocity of the balloon is zero because it is released from rest. Therefore, to determine the time it takes for the balloon to hit the ground, we only need to consider the height, which is 10 meters above the ground.
Using the equation x = x0 + v0t + 1/2at^2, where x is the position, x0 is the initial position, v0 is the initial velocity, t is the time, and a is the acceleration, we can plug in the known values and solve for t:
10 = 0 + 0*t + 1/2*(-9.8)*t^2
Simplifying the equation, we get:
t^2 - 2t = -2
Solving this quadratic equation, we find that t is approximately 2.02 seconds.
Since 2.02 seconds is greater than 1.0 second, we can conclude that the balloon will hit the ground after 1.0 s of falling.
To determine whether the balloon would hit the ground before or after 1.0 second of falling, we need to calculate the time it would take for the balloon to reach the ground using the given equations.
Using the equation x = x0 + v0t + 1/2at^2, we can set the initial position (x0) as 0 meters and the acceleration (a) as -9.8 m/s^2 (negative because it is downward), and the equation becomes x = 0 + 0 + 1/2 * -9.8 * t^2.
Since the window height (x) is 10 meters, we can rearrange the equation to solve for time (t): 10 = 1/2 * -9.8 * t^2.
Simplifying, we get -4.9t^2 = -10. Dividing both sides by -4.9 gives us t^2 = 2.04. Taking the square root, we find that t ≈ 1.43 seconds.
Since the calculated time is greater than 1.0 second, we can conclude that the balloon will hit the ground after 1.0 second of falling.
Therefore, the equation used to decide this was x = x0 + v0t + 1/2at^2, and the comparison made was checking if the calculated time was greater or less than 1.0 second.