physics problem?

Question

A ball is dropped from rest from the top of a building and strikes the ground with a speed of Vf. From ground level a second ball is thrown straight upward at the same instant that the first ball is dropped. The initial speed of the second ball is v0=vf, the same speed with which the first ball will eventually strike the ground. Ignoring air resistance, decide whether the balls cross paths at half the height of the building , above the halfway point, or below the halfway point. Give your reasoning

Answer 1

If you think about it the first ball is at the top and is stationary, then is dropped and gets to the bottom with maximum speed (terminal velocity). The second ball starts at the bottom with the speed that the first ends with and when it reaches the top of the throw it will be stationary. As both balls are being accellerated downwards by gravity they both have the same accelleration. The second ball is therefore travelling the same path as the first ball but the pattern of it's speed is the opposite.

Therefore they will cross at the halfway point where their velocities will be equal in magnitude but opposite in direction.

Answer 2

The ball is dropped, not thrown....so the speed at which it hit the ground at that instant is it's terminal velocity(the ball went from zero to vf in that distance with the help of gravity)...We are ignoring wind resistance, however we are not ignoring gravity, so the second ball leaving the ground point at the same terminal velocity(v0) at exactly the same time, should reach the height of the exact point of where the first ball was dropped from; at which point the second balls speed should be zero(the starting speed of the first ball..because it was dropped not thrown). Therefore they should cross paths at exactly the halfway point, at exactly the same speed.

Answer 3

about three-quarters up would be my guess. Because the thrown ball starts at a greater speed than the dropped ball, so it crosses the first half faster. They'll be going the same speed when they cross paths, and the dropped ball will be going faster after that. I can't show three-quarters mathematically, but that seems like a good guess.

Answer 4

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Suppose total height is H and the balls cross at x above the ground or H-x below the top

vf= sq.rt 2gH

H-x=(1/2)gt^2-------(1)

x = vf t-(1/2)gt^2----(2)

From(1) and (2)

H=vf t

vf =sq rt 2gH

t =(sqrt 2(H-x)/g )

H= [ sqrt 2gH ](sqrt2(H-x)/g)

squaring

H^2 =4H^2 - 4Hx

cancel H,

4x =3H

x =3H /4

The balls will cross at height (3H /4) that is ABOVE THE HALF WAY
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Answer 5

Above. Because the bottom part of the trajectory is faster.

Answer 6

they ll meet half way thro... reason: symmetric relation of velocities , distances n mass....