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Need help with the sign of g in free fall acceleration 
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#1
Jan1409, 06:54 PM

P: 61

1. The problem statement, all variables and given/known data
Heather and Jerry are standing on a bridge 50\;m above a river. Heather throws a rock straight down with a speed of 20\;m/s. Jerry, at exactly the same instant of time, throws a rock straight up with the same speed. Ignore air resistance. How much time elapses between the first splash and the second splash? 4.08 s 2. Relevant equations 3. The attempt at a solution i already have the answer of 4.08 s, which according to mastering physics, is correct. question is: in solving the equations i was instructed to use a positive g for heather and a negative g for jerry........i don't get this. maybe i did the problem wrong and got the correct answer. 


#2
Jan1409, 07:13 PM

HW Helper
P: 3,531

This question will use the equation:
[tex]s=ut+\frac{1}{2}at^2[/tex] where: s=displacement u=intial velocity a=acceleration due to gravity t=time elapsed If we label Heather's stone as A, and Jerry's as B, then [tex]t_{B}t_{A}=t_{elapsed}[/tex] If we substitute all our known results into the formula for stone A, we get: [tex]50=20t+\frac{1}{2}gt^2[/tex] (g=gravity) notice how all values are positive, because the displacement is downwards, the velocity of the rock is directed downwards, and the gravity is acting downwards. Similarly, all values could be negative in the equation if we take up to be the positive. Now all you have is a quadratic equation in t. For rock B, we get: [tex]50=20t+\frac{1}{2}gt^2[/tex] notice how the only difference now is that the initial velocity is negative, since the rock is being thrown upwards, or away from the displacement and gravity direction. Solve this for t as well and you can then find your difference. If you are hopeful you know what you're doing, I suggest when using the quadratic formula (or any other method to solve for t) that you leave g as a variable, and once you get your answer in terms of g, then substitute the 9.8ms^{2}. 


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