To determine whether the balloon would hit the ground before 1.0 seconds, I used the first equation of motion: \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \). In this equation, \( x_0 = 10 \) m (the initial height), \( v_0 = 0 \) m/s (the initial velocity), and \( a = -9.8 \) m/s² (the acceleration due to gravity). After substituting \( t = 1.0 \) s, we find that \( x = 10 + 0(1.0) + \frac{1}{2} (-9.8)(1.0^2) = 10 - 4.9 = 5.1 \) m ). Since the position calculated is still above the ground, the balloon has not yet hit the ground after 1.0 seconds, indicating that the time to reach the ground will be greater than 1.0 seconds.
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.
Ten meters is the balloon's starting location; its acceleration is -9.8 m/s², and its velocity is 0 m/s. The balloon has not yet touched down because its position is still above the earth after 1.0 seconds. As a result, the time to reach the ground will be greater than 1.0 seconds.
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Which equation did I use?
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