Your original equation is y=y(init)+v(init)t-(1/2) g t^{2}
At the final position, y=0. g=9.8 m/s^{2}, v(init)=-5 m/s,
So the equation is
y(init)-5*2 -(9.81/2)(4)=0 or y(init)-[5*2 +(9.81/2)(4)]=0
You subtract (10+9.81*2) from y(init) and you get zero. What is y (init)?
When solving an equation, we isolate the unknown at one side, by adding everything else to both sides or subtracting everything else form both sides, or multiplying/dividing both sides...
When working with free falling objects... In general... Is acceleration always calculated to be -9.80 in the equation?????
because I origninally thought a = -g meaning -(1/2) (-g) t^2 = (1/2) (g) t^2
... but I see that the original equation already has a neg sign there for acc... "-" (1/2) g t^2.
No, the acceleration due to gravity is always downward so if you take + in the upward direction, g is negative. It is the initial velocity that is positive if the object is thrown upward, negative if it is thrown downward, 0 if it is "dropped".
Yes, the equation is written assuming the upward direction is positive.
The original equation you learn for constant acceleration a is[tex]y=y_0+{v_y}_0 t+\frac{1}{2}at^2[/tex]If you take the upward direction to be positive, you have a=-g=-9.8 m/s^{2}, so you get[tex]y=y_0+{v_y}_0 t-\frac{1}{2}gt^2[/tex]This is the equation you're using, so by using it, you've assumed the upward direction is positive. On the other hand, if you take the downward direction to be positive, you would have a=+g=+9.8 m/s^{2}, so you get[tex]y=y_0+{v_y}_0 t+\frac{1}{2}gt^2[/tex]In both cases, g=+9.8 m/s^{2} since g is the magnitude of the acceleration due to gravity.
As ehild said, both ways will work, but you have to be consistent and adjust the signs on the various quantities depending on which convention you're following.
Suppose a body moves with uniform acceleration. Its position in terms of time is y(t)=y(init) + v(init) t+a/2 t^{2}.
In case gravity is the only force acting on the body, you know that it accelerates downward, and the magnitude of the acceleration is g=9.8 m/s^{2}. The value of g can be given with more or less accuracy, and it is different at different places of Earth, but it is always a positive number.
Before you start to solve a problem you draw a figure and show what direction you mean positive. You should choose the sign of both the acceleration and initial velocity with respect to this direction.
In the problem, y(init) is positive, and the position of the stone is always positive till it reaches the ground, where y is defined zero.
The initial velocity is downward, so it is negative, -5 m/s. Have been the stone thrown upward, the initial velocity would be positive.
The acceleration is downward, and its magnitude is equal to g, so a=-g =-9.8 m/s^{2}. So your equation is