Kinematics balloon problem help

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SUMMARY

The discussion focuses on a physics problem involving kinematics, specifically the trajectory of a water balloon dropped from a height of 18.0 meters. The calculations show that the balloon takes approximately 1.82 seconds to reach the ground, while the professor, walking at a speed of 0.450 m/s, takes about 2.22 seconds to reach the point directly beneath the window. Therefore, the balloon will hit the ground before the professor arrives, confirming that it does not hit his head. The calculations utilize the equations of motion under constant acceleration due to gravity.

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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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