# Need help with the sign of g in free fall acceleration

by hachi_roku
Tags: acceleration, fall, free, sign
 HW Helper P: 3,531 This question will use the equation: $$s=ut+\frac{1}{2}at^2$$ 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 $$t_{B}-t_{A}=t_{elapsed}$$ If we substitute all our known results into the formula for stone A, we get: $$50=20t+\frac{1}{2}gt^2$$ (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: $$50=-20t+\frac{1}{2}gt^2$$ 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.