Answer:
6.17 m
Explanation:
We are given that
Initial speed of rubber ball, u=12 m/s
Total height, h=7 m
Initial speed of second ball, u'=0
We have to find the height at which the two balls collide.
Let first rubber ball and second ball strikes after t time.
For first ball
Distance traveled by first ball in time t
![S=ut-\frac{1}{2}gt^2](https://tex.z-dn.net/?f=S%3Dut-%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
Substitute the value
![S=12t-\frac{1}{2}(9.8)t^2](https://tex.z-dn.net/?f=S%3D12t-%5Cfrac%7B1%7D%7B2%7D%289.8%29t%5E2)
...(1)
Distance traveled by second ball in time t
![7-S=\frac{1}{2}(9.8)t^2](https://tex.z-dn.net/?f=7-S%3D%5Cfrac%7B1%7D%7B2%7D%289.8%29t%5E2)
.....(2)
Using equation (2) in equation (1) we get
![S=12t-(7-S)](https://tex.z-dn.net/?f=S%3D12t-%287-S%29)
![S=12t-7+S](https://tex.z-dn.net/?f=S%3D12t-7%2BS)
![\implies 12t=7](https://tex.z-dn.net/?f=%5Cimplies%2012t%3D7)
sec
Now, using the value of t
![S=12(\frac{12}{7})-4.9(\frac{12}{7})^2](https://tex.z-dn.net/?f=S%3D12%28%5Cfrac%7B12%7D%7B7%7D%29-4.9%28%5Cfrac%7B12%7D%7B7%7D%29%5E2)
S=6.17 m
Hence, at height 6.17 m the two balls collide .
It's really difficult to make out the circuit above. Quite frankly, your
question leaves me to wonder how far 'above' it may be.
The best I can do will be to try and fabricate an answer based on the
information given in the text of the question, augmented only by my own
training, chutzpah, and life experiences.
If the circuit ... wherever it is ... consists entirely of the single 3-ohm
resistance and no other components, and the current through the
resistance is 10 Amperes, then
Voltage = (Current) x (resistance) = 30 volts .
A una distncia de 12.0 m dsde una Fuentes puntual