Find the time it takes for a water balloon dropped from a second story window and accelerating at 9.8 m/s² downward to hit a passerby on the sidewalk below. The balloon starts from rest and hits with a velocity of 15.9m/s
To find the time it takes for the water balloon to hit the passerby, we can use the equations of motion. The equation that relates distance (d), initial velocity (u), acceleration (a), and time (t) is:
d = ut + (1/2)at²
In this case, the initial velocity (u) is 0 m/s because the balloon starts from rest. The acceleration (a) is -9.8 m/s² because the balloon is accelerating downward due to gravity. The distance (d) is not given directly, but we can assume it is the height of the second-story window.
We can rearrange the equation to solve for time (t):
d = ut + (1/2)at²
d = 0t + (1/2)(-9.8)t²
d = -(4.9)t²
Now, substitute the given values into the equation. The final velocity when the balloon hits the passerby (v) is 15.9 m/s. We can use the equation:
v = u + at
15.9 = 0 + (-9.8)t
15.9 = -9.8t
Now, solve for t:
t = -15.9 / -9.8
t ≈ 1.63 seconds
Therefore, it takes approximately 1.63 seconds for the water balloon to hit the passerby on the sidewalk below.