Suppose you throw two identical stones from the top of a tall 200m building. You

Question

Suppose you throw two identical stones from the top of a tall 200m building. You throw one directly upward with a speed of 10 m/s and the other directly downward with the same speed.

Part (a): When they hit the street below, both will be traveling at the same speed. Why?

Part (b): What will that speed be?

Part (c): Suppose simultaneously one stone is thrown straight up from the base of another building (street level) and the other straight down from the top of the building (roof level) with the same speed of 10 m/s. Assume the height of the building is 6m; find the location above the base of the building of the point where the stones cross.

Explanation / Answer

a)Suppose you just drop one of the stones. One second later it is traveling downwards at about 10m/s (9.8m/s actual). At that point where it is traveling at 10 m/s you catch it. Now you throw it upwards at 10 m/s. It will go up to where it was originally dropped. Then it drops back down and again it reaches 10 m/s at the point where you tossed it up. You could do that all day long. Tossing it up and catching it. And as along as you tossed it upwards at 10 m/s, it would always go to the same height. And it would always come back down with the same velocity.

So, now, instead of just dropping a stone from the side of a building, you throw it downwards at 10 m/s. And you then toss one upwards at 10 m/s. The one that is tossed upwards comes back down again and passes the spot you threw it upwards at 10 m/s (heading down). That is the same velocity that you threw the one downwards. So they hit the ground with the same energy at the same velocity.

b) v^2 = u^2 +2gs = 10^2 + 2 *9.8 * 200 = 4020 m/s^2

c) v^2 -10^2 = 19.6 *x

v^2 -10^2 = 19.6 (6-x)

19.6(6-x) = 19.6 x

2x= 6

x= 3m

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