Ball velocity after 0.5 seconds with momentum principle

[jmmcarync]

Ball is kicked from location <6,0,-9> with initial velocity <-10,17,-6>. The ball's speed is low enough so that air resistance in negligible.

What is the velocity of the ball after .5 seconds of being kicked?

Use Momentum Principle


I know that the x and z components will stay the same, but I do not understand how we can find the velocity if they do not give the mass. I keep coming up with

$$_{V}iy$$+(1/2)*($$_{F}y$$/m)*($$\Delta$$t)

What am I missing?

 

[chaoseverlasting]

Here, the only force in the y direction is gravity, and f/m=g.

 

[jmmcarync]

Alright so f/m=g.

V(yfinal)=$$_{V}iy$$+(1/2)*(g)*($$\Delta$$t)
=<17>+[(1/2)(-9.80)]*.5
=<17>+(-2.45)
=14.55 m/s

This answer is still incorrect. what's going on?

 

[rootX]

I think F is acting in the z direction.
and
v(x,y,z) is = to <-10,17,..>+t<0,0,-g>

 

[jmmcarync]

I don't think so, my values for the x and z componenets are correct. Like chaos said, gravity is the only force acting on the ball, so that would be in the y direction.

 

[rootX]

lol, you are using wrong formula

vf = v0 + at

because time is also 1/2, so i din't catch it. sorry :shy:

 

[jmmcarync]

aha! you got it rootx. good job thanks much.