Correct. g = -9.8ms^-2It's because in projectile motion treatment, the positive value of displacement is assumed to be "up", and accordingly all derivatives have the same assumption embedded in them. V is "up" and -V is "down". a is "up" and -a is "down".
On the surface of Earth, the magnitude of the acceleration vector is assumed to be g, and it's direction "-", or "down", i.e., towards the negative values.
You can use proper vector notation and get the same results. It's just that most textbooks tackle projectile motion before properly explaining vectors.
I was referring to thisAs long as you use a reference frame in which the displacement "upwards" is in the positve direction, you have to use -g, since that's the direction towards which the gravity acts.
As far as I can see from your posts in the other thread I think you're refering to, the problem is that you use the negative value for g twice, so to speak, and thus get the wrong result.
The general equation for projectile motion is: [itex]r=\frac{at^2}{2}+V_0t+r_0[/itex], where r, a, and V are vectors.
If you reduce the equation to one dimension(y; where the positive values are "up"), you can simply say that [itex]a_y=-g[/itex], you end up with [itex]y=\frac{-gt^2}{2}+V_{y0}t+y_0[/itex]
You seem to be wrongly using [itex]r=\frac{-at^2}{2}+V_0t+r_0[/itex]
and then substituting [itex]a_y=-g[/itex], which erroneously gives you [itex]y=\frac{gt^2}{2}+V_{y0}t+y_0[/itex]
That's all there is to it.
The value of g is not -9.8. It's 9.8. If you're asked to substitute the value of the scalar g into your equation, you substitute the value without any information about the direction of the vector(i.e., the '-' sign). If you see g in an equation, it's a safe bet the direction was already included at one time.I was referring to this
https://www.physicsforums.com/showthread.php?t=730711
If my value for g = -9.8, I get a correct value but in the negative. The book has a positive value. This means that, and echoed by another poster, that g must be positive.
The confusion is, why should g in this equation be a magnitude. It doesn't make an iota of sense.
You are correct.The value of g is not -9.8. It's 9.8. If you're asked to substitute the value of the scalar g into your equation, you substitute the value without any information about the direction of the vector(i.e., the '-' sign). If you see g in an equation, it's a safe bet the direction was already included at one time.
If you were asked to substitute some value for the vector ##a_y## then you'd have to put the minus together with the magnitude(g), as you'd be asked to provide the full information.
The equation you're asked to use: [itex]x = \frac{v_0^2}{g}sin{2Θ}[/itex]
is derived from:
[tex]y=\frac{-gt^2}{2}+V_{y_0}t+y_0[/tex]
and
[tex]x=V_{x_0}t+x_0[/tex]
the minus sign was already included when we substituted ##a_y=-g## when we reduced the general vector equation to one dimension.
Try solving the first one for t, then substitute the result to the second one, and you'll see what happened to the minus sign that was originally there.