Kinematics balloon problem help

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A student analyzes a physics problem involving a water balloon dropped from 18.0 m above the ground as a professor walks underneath. The calculations show that the balloon takes approximately 1.82 seconds to reach the ground, while the professor takes about 2.22 seconds to reach the point directly beneath the window. Since the balloon reaches the ground before the professor arrives, it will not hit his head. The student's calculations appear correct, confirming that the balloon drops before the professor gets there. The discussion emphasizes understanding kinematic equations in real-world scenarios.
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Homework Statement



A student at a window on the second floor of a dorm sees his physics professor coming along the walkway 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. If the prof is 180 cm tall and walks at a rate of 0.450 m/s, does the balloon hit his head?

The Attempt at a Solution



Yo = 0
y = -18.0 m + 1.8 m = -16.2 m
g = -9.81 m/s^2
Vo = 0
X = 1.00 m
Vxo = 0.450 m/s

Balloon Drop:

y = yo + Vo*t - 1/2*g*t^2
-16.2 m = 0 + 0 - 1/2(9.81 m/s^2)t^2
t^2 = 16.2 m/4.9 m/s^2
t^2 = 3.31 s^2
therefore, t = sqrt3.31 s^2
= 1.82 s

Professor Walk:

x = Vxo*t
1.00 m = (0.450 m/s)t
t = 1.00 m/(0.450 m/s)
=2.22 s

Therefore, the balloon drops before the professor gets there.

Can anybody please tell me if my attempt is o.k.?
 
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Looks fine to me.
 


Thanks Kurdt!
 

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