Hamad -
Well, I don't want to do your homework for you, but I can help to
show you how this problem works... I think the easiest way to approach
this is in terms of energy, namely potential energy. The thing about
energy is that it's always conserved. That is, even though it may get
transfered from one object to another, the total amount is always the
same.
In this problem, you've really got two different spots in time to
think about. The first one has the first acrobat in the air, just about
to start falling. But he's not falling
yet, so all he has is
potential energy. The second point is when the second acrobat is in the
air as high as he's going to get - he's halfway in between moving
upwards and moving downwards, so he's not actually moving at all. This
means that all he's got is potential energy.
The equation for (gravitational) potential energy is
mass*gravity*height. So the energy at first is equal to the first
acrobat's mass, times the acceleration of gravity (9.81m/s^2), times
how high up he is. The energy at the end is equal to the second
acrobat's mass, times gravity, times how high up
he is. Your
problem makes it a little bit easier, since it gives you the acrobats'
weights instead of masses. The weight equals the mass times the
acceleration of gravity, so you get two equations for the potential
energy at two different times, and you know that both have to be equal:
Energy at first = Acrobat 1's Weight * Acrobat 1's Height
Energy at last = Acrobat 2's Weight * Acrobat 2's Height
Now you can just set these two equations equal to each other and solve to see how high up the second acrobat can go.
-Tamara
(published on 10/22/2007)
