Physics question below, I've worked on it and can not find the Y velocity?

Question

A ball is kicked from a location < 6, 0, -7 > (on the ground) with initial velocity < -9, 19, -4 > m/s. The ball's speed is low enough that air resistance is negligible.

What is the velocity of the ball 0.4 seconds after being kicked? (Use the Momentum Principle!)

I've got the x and z velocity's they are simple.

I would like an example of how to work this out, but if you feel giving me an answer is cheating I'll happily take info on how to use the momentum principal.

Answer 1

In the y direction its velocity ,after 0.4 second is

19 - gt where g is the acceleration due to gravity.

v(y) = 19 - 9.8x0.4 = 14.08m/s

Compute the resultant velocity by the square root of the sum of the squares of the velocities in x , y and z direction.

Answer 2

I usually take z to be up and down but since you say this is on the ground and z is -7 I guess y is up and down. Using momentum you say?

Initial momentum is: p0 = mv0
At any time > 0 we have: p = mv

The force of gravity acting on the ball is:
f = mg where g = -9.8 m/s
And f = dp/dt
mg = dp/dt

dp/dt = (p - p0) = (mv - mv0)/t = m(v - v0)/t
gt = v - v0
v = v0 + gt
v = 19 - 9.8(0.4) = 15.08 m/s