Finding time when you have initial velocity and position

[oliampian]

***Sorry I meant to say finding time when you have initial velocity and position***
1. Homework Statement
A test rocket is launched by accelerating it along a 200.0-m incline at 1.36 m/s2 starting from rest at point A (the figure (Figure 1) .) The incline rises at 35.0 ∘ above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored).

Find the maximum height above the ground that the rocket reaches.
Find the greatest horizontal range of the rocket beyond point A.

Known:
V_o = 0 (because starting from rest)
a_{on incline} = 1.36 m/s²
a_{off incline} = -9.8 m/s²
ramp distance = 200m
θ = 35°

Homework Equations

x = x_o + V_ot + (0.5)at²
V = V_o + at
V² = V_o² + 2aΔx

The Attempt at a Solution

I already found the first part, which turned out to be 124m when rounded to three sig figs.
I found through calculations that V_{at end of incline} = 23.323m/s:
using the V² = V_o² + 2aΔx
V_{at end of incline} = √(2(1.36m/s²)(200m))
V_{at end of incline(x-component)} = 23.323m/s(sin(35)) = 19.1050m/s
Distance in x-component = D_x = 200m(cos35) = 163.8304

So to find the total horizontal distance so far I have the equation:
x = 163.8304 + 19.105t - 0.5(9.8)t²

What I'm stuck on is trying to find the total time so I can plug that into the equation.

 

[SteamKing]

***Sorry I meant to say finding time when you have initial velocity and position***
1. Homework Statement
A test rocket is launched by accelerating it along a 200.0-m incline at 1.36 m/s2 starting from rest at point A (the figure (Figure 1) .) The incline rises at 35.0 ∘ above the horizontal, and at the instant the rocket leaves it, its engines turn off and it is subject only to gravity (air resistance can be ignored).

Find the maximum height above the ground that the rocket reaches.
Find the greatest horizontal range of the rocket beyond point A.

Known:
V_o = 0 (because starting from rest)
a_{on incline} = 1.36 m/s²
a_{off incline} = -9.8 m/s²
ramp distance = 200m
θ = 35°

Homework Equations

x = x_o + V_ot + (0.5)at²
V = V_o + at
V² = V_o² + 2aΔx

The Attempt at a Solution

I already found the first part, which turned out to be 124m when rounded to three sig figs.
I found through calculations that V_{at end of incline} = 23.323m/s:
using the V² = V_o² + 2aΔx
V_{at end of incline} = √(2(1.36m/s²)(200m))
V_{at end of incline(x-component)} = 23.323m/s(sin(35)) = 19.1050m/s
Distance in x-component = D_x = 200m(cos35) = 163.8304

So to find the total horizontal distance so far I have the equation:
x = 163.8304 + 19.105t - 0.5(9.8)t²

Where does this equation come from? And when did gravity start acting sideways?

When the rocket leaves the ramp, the engines turn off, so the rocket just coasts until it lands.

What I'm stuck on is trying to find the total time so I can plug that into the equation.

When the rocket leaves the ramp, it acts like any other projectile. You have an initial velocity and an angle at which the rocket is traveling relative to the horizon.