A ball is kicked from a location < 7, 0, -6 > (on the ground) with initial velocity < -9, 15, -3 > 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 divide 7,0,6 by .4 and got an answer of <17.5, 0 , -15>, but it was wrong. What am I leaving out?

I'm assuming that the 'x' axis is like a right left direction, the 'y' axis like a vertical direction and the 'z' axis like an in-out of paper direction.

From \(7\\0\\-6\), the ball is kicked out of the paper, with velocity [math]\(-9\\15\\-3\)

If I make the assumptions I made before, then the vertical component is 15.

Using the motion formula v = u+at, in 0.4 seconds, the final vertical velocity will be:

v = (15) - (9.81)(0.4) = 11.076 (acceleration due to gravity is 9.81m/s^2

Then, the total final velocity will be: \(-9\\11\\-3\)

(all to the nearest integer)