Basic Physics

I have a rather simple problem.
Earlier today I asked for help concerning how to calculate the deceleration speed of an ascending object traveling at 150 km/h before it stops and weighing 360 kilos.
I got this:
Calculation: Velocity = 41.666667 Gravity = 9.81
        S = Vot+1/2at^
        S = 41.666667(4.2517) + 1/2(-9.8)(4.2517^2)
        S = 177.1541681+ -88.57706916
        S = 88.48671416 ≈ 88.5

My question is; what is (4.2517)? Where does it come from?

>>1
>My question is; what is (4.2517)? Where does it come from?

1) A number
2) Either the top row of keys or the number pad.

deceleration speed... before it stops
What does that even mean?  Deceleration is constant, and = 9.8 (acceleration = -9.8).  Speed just before it stops is 0.  You want the rate of change of acceleration?  That's also 0.

4.2517 is how long it takes for the particle to stop.  The S is how high up it goes.

>>2
lol

Op here

Yes, it is constant. Ok, the real question is; how high will this object go before it starts falling again after having traveled at 150 km/h for 5 seconds (reaching 208.5 meters above ground).
Here, the "engine" or whatever, is turned off. What I need is the equation to calculate this.
I was wondering what 4.2517 was, since, well, how did the person who calculate figure out the time it took for the objects speed to reach 0.