Astudent at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she drops a water balloon from 18.0 meters above the ground when the professor is 1.00 meter from the point directly beneath the window. if the professor is 1.70 meters tall and walks at a rate of 0.450 m/s, does the balloon hit her? if not, how close does it come? (10 pts)

Refer to the diagram shown below.

In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s

The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m

Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.

The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s

The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.

 
The balloon misses the professor, and falls 0.175 m in front of the professor.