Particle thrown from earth's surface

[barz1]

Homework Statement

Okay so, a particle is thrown upward from the surface of the Earth with an initial velocity of 4 km/s. Force of gravity equals 9.81 m/s^2.
Find max height the particle will reach.

Homework Equations

Since the particle remains close enough to the earth, I believe simple constant acceleration equations apply.
v_f=v_i + a*t I used to solve for time
Y_f=v_i + 1/2a*t² + Y_o I used to solve for height above earth

The Attempt at a Solution

So I feel like this should be the easiest problem in the world. The question tries to throw you off by giving velocity in km/s and acceleration in m/s so converting accel. to km/s since they want an answer in km. I solved for t and got 407.75s then I just plugged this into my second equation and solved for Y_f, or height above the earth.
Buttt I am not getting the right answer, I didn't try adding my Y_f to the radius of the Earth (although I don't really know why I would be expected to) and this question is from a chapter regarding universal gravitation. So I guess I am just wondering did I go about this problem the right way or no?

Thanks!

 

[gmsss]

use equation v²-u²=2as
use sign convention carefully

i think you know that at max height velocity is zero

 

[nagaromo]

Btw the ACCELERATION due to gravity equals 9.81m/s^2.
So using the equation Vf^2 = Vi^2 + 2ad you can find the max height by solving for d.
Here's a clue: What is the final velocity at maximum height?