Sunday

April 20, 2014

April 20, 2014

Posted by **Genevieve** on Saturday, July 29, 2006 at 11:19pm.

thank you very much

The teacher's head is H = 18.0 - 1.7 = 16.3 m below the window from which the water balloon is dropped. The time required to fall that distance is

t = sqrt(2H/g) = 1.82 s.

The teacher moves horizontally

t*0.45 m/s = 0.82 m in that time. If he was 1.0 m behind the impact point when it was dropped, he will be 0.18 m behind the impact point when the water balloon reaches the height of his head.

That's pretty close. You need to consider the finite size of the water balloon and the person before deciding on an answer. The teacher will have moved even closer to the balloon before it hits the ground.

I would say the water balloon would barely miss his head but probably hit his forward-extended leg before it hits the ground.

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