In the figure below, a student at a window on the second floor of a dorm sees his math professor walking on the sidewalk beside the building. He drops a water balloon from 18.0 m above the ground when the prof is 1.00 m from the point directly beneath the window. The professor is 170 cm tall and walks at a rate of 0.450 m/s. How close does the balloon come to hitting the professor? (Enter 0 if the balloon hits the professor.)

To determine how close the balloon comes to hitting the professor, we need to calculate the horizontal distance traveled by the balloon before it hits the ground.

First, let's determine the time it takes for the balloon to hit the ground. We can use the formula for free-falling objects:

h = (1/2)gt^2

Where:
h = initial height of the balloon (18.0 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time taken

Rearranging the equation to solve for time, we get:

t = sqrt(2h / g)

Substituting the values, we have:

t = sqrt(2 * 18.0 / 9.8) = 2.14 seconds (approx.)

Next, let's determine how far the professor walks during this time. The distance the professor walks can be calculated using the equation:

distance = speed * time

Where:
speed = rate at which the professor walks (0.450 m/s)
time = 2.14 seconds

Substituting the values, we have:

distance = 0.450 * 2.14 = 0.963 meters (approx.)

Since the professor walks 0.963 meters during the time the balloon falls, the balloon comes within 0.963 meters of hitting the professor.