Velocity/Speed Problems (Studying for exam)

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Homework Statement


A person throws a ball with 13,5 m/s straight up. Calculate the maximum height that it will reach from the point where it was thrown. Also find the velocity when the ball is 1.50m above the starting point.

Sorry if the problem is hard to understand, had to translate it from spanish, hopefully it's clear enough.

Homework Equations


Vx = V0x + axt
x = x0 + v0xt + 1/2 ax t^2

The Attempt at a Solution


13.5m/s = 0 + 9.81 * t
13.5/9.81 = t
t = 1.38 s (what it took to reach the highest point)

x = 0 + 13.5 * 1.38 + 1/2 * -9.81 * 1.37^2
x = 9.42 m (maximum height?)

Not sure if the problem is correct, if it's not any guidance is appreciated. I'm also missing what's the velocity 1.50m above the starting point.
 
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t is correct (1.376 s), but you are using 1.38 and 1.37^2 in the same equation. Hang on to the couple extra digits until you have the final result.

You know x0, x, Vx0, and a. There is a kinematic equation that you have not posted that will allow you to find vx directly.