Deceleration and time

[Quinch]

Hello everyone. While this isn't for homework perse, I've got a simple question that has been frustratingly difficult to dredge out the answer/formulae to.

 

How can I calculate the time it takes an object to rise and fall from elevation A to elevation B at initial upward speed of X units per second and gravity of Y units per second?

 

My Google-fu has failed me, so I'd be appreciative of a working method for this.

 

{Edit: And I've apparently posted this in the wrong forum. Go me. Blah.}

[swansont]

[math]s = {v_0}t + \frac{1}{2}at^2[/math]

 

s is displacement, v is velocity, a is acceleration. You keep track of direction with + and - values.

[Bignose]

Just to echo what swansot said, but just in a different way, a deceleration is an acceleration with just the opposite sign. The math works out either way.

[alwynj48]

2X/Y=T

 

which is just a short way of saying if its accelerating at Y m/s/s and the initial velocity is X m/s it will take X/Y to decelerate to 0 and the same time to accelerate back to X m/s.

 

or say its starts at 100 m per s every second it decreases by 10m per s so it would take 10 s to stop and 10 s to fall back down.

 

Don't forget what acceleration is ie the change in velocity per second

 

If you want to calculate the distance covered you should know that the velocity profile is triangular so the average velocity is half the peak velocity. So multyply the time by half the velocity and give you the distance or in the case the distance travelled.