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.)

18 Skdjek

To find out how close the balloon comes to hitting the professor, we need to find out the horizontal distance the balloon travels as it falls from the window to the ground.

First, let's calculate the time it takes for the balloon to fall from the window to the ground.

Given:
Initial height (h) = 18.0 m
Acceleration due to gravity (g) = 9.8 m/s^2

Using the kinematic equation for vertical motion:
h = 0.5 * g * t^2

Rearranging the equation to solve for time (t):
t^2 = (2 * h) / g

Substituting the given values:
t^2 = (2 * 18.0) / 9.8
t^2 = 3.673

Now, take the square root of both sides to find t:
t ≈ √3.673
t ≈ 1.92 seconds

Therefore, it takes approximately 1.92 seconds for the balloon to fall from the window to the ground.

Next, we need to calculate the horizontal distance the professor walks during this time.

Given:
Professor's walking speed (v) = 0.450 m/s
Time (t) = 1.92 seconds

Using the formula distance (d) = speed (v) * time (t):
d = v * t
d = 0.450 * 1.92
d ≈ 0.864 meters

Therefore, the balloon comes approximately 0.864 meters close to hitting the professor.