g does not change
velocity changes, but g does not.
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Sign of g in these equations
Hi,
I'm trying to come up with an equation to describe that relates the distance a ball is dropped from a height of H that hits a plate at a 45 degree angle and then goes off on a parabolic path to a distance of R (basically Height between ball and bounce plate vs. range). I'm a little confused for what the sign of g (the ball's acceleration) would be as the ball goes through its motion. I figured that when its dropped to hit the plate the sign is negative, but when its in the parabolic path is it positive?
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Answers
Its just that for my equation I came up with the ball having a velocity of sqrt(2(-9.8)*(height between plate and where ball dropped) and the time it takes to reach the particular range as being sqrt((2*distance between bounce plate and the ground)/9.8). I assumed that the bounce plate was the "origin" so the first height in the velocity equation would be positive and the second height in the time equation in the time equation would be negative. If I make both 9.8s negative and then multiply v*t the quantity under the sqrt is negative. So I'm really confused about sign change. Please help!
Is this in reference to http:/Q_23746570.html ?
If so, which direction is positive H and h?
If you used different signs for H and h, that would cause g to change signs too.
For the first part: mgH = ½mv² where v is velocity just before the bounce
==========================
For the second part: ½gt² = h ==> t = sqrt(2h/g)
==> v = R/t = R*sqrt(g/2h)
==> E = ½mR²g/2h
where t is the time between the the bounce and the ground
v is the velocity just after the bounce
E is kinetic energy just after the bounce
if you're using V^2 = U^2 + 2gs (with U = 0 here) , then you need to have g positive if s is positive, if you were 100m from a finish line at the start of a race, and you accelerated at -1m/s/s then you would never reach the finish line, hence the equation having no real solution.
If the point where the ball is dropped is +100 and the point where it hits is 0, then you have covered a distance of -100m and you need a negative g
so, if going up is positive, then g is negative
if going down is positive then g is positive
so I think you actual need
sqrt(2(-9.8)*(-1 * height between plate and where ball dropped)
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by: gpiero74Posted on 2008-09-22 at 04:00:22ID: 22538847
you have to assume the g acceleration always with the same sign.
It always give an acceleration versus the ground. the speed of you ball increase always going to the ground and decrease when it start to go to the sky!