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.
To determine if the balloon will hit the ground before or after 1.0 second of falling, we can use the equation x = x0 + v0t + 1/2at^2, where x is the final position, x0 is the initial position (10 meters), v0 is the initial velocity (0 m/s), and a is the acceleration due to gravity (-9.8 m/s^2).
We substitute the values into the equation:
x = 10 + (0)(1) + 1/2(-9.8)(1)^2
x = 10 - 4.9
x = 5.1 meters
Since the final position after 1.0 second is 5.1 meters, which is less than the height of the window (10 meters), the balloon would not hit the ground by then. Therefore, the balloon will hit the ground after 1.0 second of falling.
To determine whether the balloon would hit the ground before or after 1.0 s of falling, we can use the equations of motion given. The equation that we can use to solve this problem is x = x0 + v0t + (1/2)at^2. We need to calculate the time it takes for the balloon to hit the ground and check if it is less than 1.0 s. Using the formula x = x0 + v0t + (1/2)at^2, we can substitute the given values: x = 10 m, x0 = 0 m, v0 = 0 m/s, and a = 9.8 m/s^2. Solving for t, we get t = √[(2(x-x0))/a] = √[(2(10-0))/9.8]. By calculating this, we find t ≈ 1.43 s, which is greater than 1.0 s. Hence, the balloon would hit the ground after 1.0 s of falling, implying that it would not hit the ground before 1.0 s.
To determine whether the balloon will hit the ground before or after 1.0 s of falling, we need to use the equations of motion given.
Let's start with the first equation: x = x0 + v0t + (1/2)at^2
In this equation, x represents the final position of the balloon (in this case, the ground), x0 is the initial position (10 meters above the ground), v0 is the initial velocity (0 m/s since the balloon is released from rest), t is the time (1.0 s), a is the acceleration due to gravity (-9.8 m/s^2).
Plugging in these values, we get x = 10 + 0 + (1/2)(-9.8)(1.0)^2 = 10 - 4.9 = 5.1.
So, according to this equation, the balloon will have a position of 5.1 meters above the ground after 1.0 s of falling.
Since 5.1 meters is still above the ground, we can conclude that the balloon will not hit the ground before 1.0 s of falling.
Therefore, the equation x = x0 + v0t + (1/2)at^2 was used to decide that the balloon would not hit the ground by then.