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- Thread starter noodle21
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krab

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2. Anyway, your friend's ball starts upward with a velocity v, but your ball starts with velocity = 0. So whose ball gets to the halfway point first? Think about it.

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HallsofIvy

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Originally posted by noodle21

You drop a ball from a window on an upper floor of a building. it strikes the ground with velocity v. You now repeat the drop, but have a friend down on the street who throws another ball upward at velocity v. Your friend throws the ball upward at exactly the same time that you drop yours from the window. At some location, the balls pass each other. Is this location at the halfway point between window and ground, above this point, or below this point?

Originally posted by krab

1. You are in the wrong forum. This one is Quantum Mechanics. There is a forum for help with college physics.

2. Anyway, your friend's ball starts upward with a velocity v, but your ball starts with velocity = 0. So whose ball gets to the halfway point first? Think about it.

Well, apparently it got moved! In your second comment, you seem to be implying that since the ball thrown up starts with speed v while the ball dropped starts with speed 0, the ball thrown will "move faster". Don't you think acceleration (and deceleration) will have something to do with it?

Suppose the window is h meters above the ground. Both balls have an acceleration of -g. The ball dropped will have speed (negative so downward) of v1(t)= -gt and the ball thrown upward will have speed (positive so upward) v2(t)= v- gt.

The height, at time t, of the dropped ball is

h1(t)= -(g/2)t

However, v is not just some arbitrary speed. It is the speed the dropped ball has when it hits the ground: The dropped ball hits the ground when h1(t)= (-g/2)t

t= [sqrt](2h/g). At that time, v= -g([sqrt](2h/g)= -[sqrt](2hg).

Of course, the speed of the ball thrown up is [sqrt](2hg).

They will pass when they both have the same height at the same time:

-(g/2)t

By golly, the accelerations cancel out! This reduces to

[sqrt](2hg)t= h so t= h/[sqrt](2hg)= [sqrt](h/2g).

At that time, h1= (-g/2)(h/2g)+ h= -h/4+ h= (3/4)h.

Of course, h2= [sqrt](2hg)[sqrt](h/2g)-(g/2)(h/2g)

= h- h/4= (3/4)h.

Well, I'll be! krab was right! The ball thrown upward, because it had the greater initial speed (and we DIDN'T have to take acceleration into account- it canceled out) goes farther. The two balls pass 3/4 of the way up to the window.

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