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 can be used to describe its motion over time:

Question

x=x0+v0t+1/2 at^2 (the 0s are slightly under the letter in front of them)

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.

Answer 1

To determine if the water balloon would hit the ground before or after 1.0 seconds of falling, I used the first equation of motion: \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \). In this scenario, \( x_0 = 10 \) meters (the initial height), \( v_0 = 0 \) (the initial velocity when it is dropped), \( a = -9.8 \) m/s² (acceleration due to gravity acting downwards), and \( t = 1.0 \) seconds. Plugging in these values, we get:

\[ x = 10 + 0 \cdot 1 + \frac{1}{2} (-9.8)(1)^2 = 10 - 4.9 = 5.1 \text{ meters}. \]

At 1.0 seconds, the balloon would still be at 5.1 meters above the ground, meaning it has not yet reached the ground. Thus, the balloon would hit the ground after 1.0 seconds of falling.

Answer 2

Like a 9th grader, please.

Answer 3

To figure out if the water balloon will hit the ground before or after 1.0 seconds, I used the equation \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \). Here, \( x_0 \) is the starting height (10 meters), \( v_0 \) is the initial speed (0 because it’s just dropped), \( a \) is the acceleration due to gravity (-9.8 m/s²), and \( t \) is 1.0 seconds. Plugging in the numbers gives:

\[ x = 10 + 0 \cdot 1 + \frac{1}{2} (-9.8)(1)^2 = 10 - 4.9 = 5.1 \text{ meters}. \]

Since 5.1 meters is still above the ground, the balloon hasn't hit the ground yet after 1.0 seconds. This means it will hit the ground after 1.0 seconds.

Answer 4

give me a better comparision

Answer 5

Sure! To see if the water balloon will hit the ground before or after 1.0 seconds, I used the equation \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \). Here, \( x_0 = 10 \) meters (the height of the window), \( v_0 = 0 \) (the balloon starts from rest), and \( a = -9.8 , \text{m/s}^2 \) (the acceleration due to gravity).

After calculating, I found:

\[ x = 10 + 0 \cdot 1 + \frac{1}{2} (-9.8)(1)^2 = 10 - 4.9 = 5.1 \text{ meters}. \]

Since the balloon is still at 5.1 meters after 1.0 seconds, I compared this height to the ground level (0 meters). Since 5.1 meters is above ground level, it shows that the balloon would not hit the ground until after 1.0 seconds has passed. Therefore, it will take longer than 1.0 seconds for the balloon to hit the ground.