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## Homework Statement

A rocket, initially at rest, is fired vertically with an upward aceleration of 10m/s². At an altitude of .50 km, the engine of the rocket cuts off. What is the maximum altitude it achieves?

## Homework Equations

V² = V_o² + 2aΔx

t = (V-V_o)/a

## The Attempt at a Solution

Well I wanted to find the Final Velocity at the .50km first so

V² = V_o² + 2aΔx

V² = 0² + 2(10)(500m) *Converted .50km to m

V² = 10000

V = 100m/s

Since that final velocity

t = (V-V_o)/a

t = (100 - 0) / 10

t = 10seconds

So it takes 10 seconds to travel the 500m

But now I'm stuck...I see that to find the final altitude:

The X_o is now 500m

The V_o is now 100m/s

I believe that gravity -9.81m/s/s would become the new acceleration?

I'm not sure how to find out how much longer it will be in the air until it starts going down and I'm not sure how to find the final altitude.

Is this correct to this point?

EDIT: Actually the V would be 0m/s wouldn't it? I forgot that.

Now shouldn't I be able to

V = V_o + at

t = (0-100)/(-9.81)

t = 10.19367992 seconds remaining in the air so

X = V_o t + X_o + 1/2 a t²

X = 100(10.19367992) + 500 + (1/2)(-9.81)(10.19367992)²

X = 101.9367992 + 500 + (-509.6839961)

X = 92.2528031 m more than 500

So 592.2528031m

or

.59 km? No that can't be right...Okay now any help?

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