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one man drops a rock from a 100m building. At Exactly the same moment, a second man throws a rock from the bottom of the building to the top of the building. At what height do the rocks meet?
tried so many times and even checked answer ,but I could not understand.
Please help
this would never show up on mcat. However, it's still a ^ problem for elucidating the thought process.
I always use delta x= vot + etc. Using that here is bad
The key to this problem is realizing that the ball is thrown so that it will reach 100m. I didn't see this at first, however, you must think mcat so ideas like this should enter your head.
x=xo + vot + .5gt^2. for thrown rock. xo=0. So, x= vot -.5gt^2 for thrown
for dropped x=100 - .5gt^2. These distances are equal, since the x is equal to the distance the thrown one travels and is equal to the distance at which the second rock is at.
Set them equal, 100=vot vo=2000^.5 which is basically 45
100/45 is about 2.2 2.2^2 is 4.84 but you know it's less than 5. so using t=2.3, we solve for x, 100-.5(5)(10)= 75
This is a little BS because since it was translational, i didn't think I'd need energy, but it's good in a way because it lets you see the advantage of energy problems. It showed me the importance of understanding the equations.
