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:
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?
In 5 sentences explain your reasoning.
To determine whether the balloon would hit the ground before or after 1.0 second of falling, we can use the equation x = x0 + v0t + (1/2)at^2, where x is the position of the balloon, x0 is the initial position, v0 is the initial velocity, t is time, and a is the acceleration due to gravity.
Since the balloon is initially at a height of 10 meters (x0 = 10) and the acceleration due to gravity is 9.8 m/s^2 (a = 9.8), we need to find the time at which the balloon reaches the ground.
We can set x to 0, as the balloon hits the ground at that point, and solve the equation for t. Plugging in the values, we get 0 = 10 + 0 - (1/2)(9.8)(t^2).
Simplifying the equation, we have 0 = 10 - 4.9t^2. Rearranging the terms, we get t^2 = 10/4.9.
We can take the square root of both sides to find t, which is approximately 1.42 seconds. Since 1.42 is greater than 1.0 second, the balloon would hit the ground after 1.0 second of falling.