Calculating Ball Velocity with Momentum Principle: Newb Question Answered

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The discussion revolves around calculating the ball's velocity after 0.5 seconds using the momentum principle. The initial velocity components are given, and it's noted that air resistance is negligible. Participants clarify that only gravity affects the y-component, leading to the correct application of the formula for final velocity. The confusion primarily stems from misapplying the formula and misunderstanding the role of time in the calculations. Ultimately, the correct approach confirms that the x and z components remain unchanged while gravity influences the y component.
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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)*(\Deltat)

What am I missing?
 
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Here, the only force in the y direction is gravity, and f/m=g.
 
Alright so f/m=g.

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

This answer is still incorrect. what's going on?
 
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I think F is acting in the z direction.
and
v(x,y,z) is = to <-10,17,..>+t<0,0,-g>
 
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.
 
lol, you are using wrong formula

vf = v0 + at

because time is also 1/2, so i din't catch it. sorry :shy:
 
aha! you got it rootx. good job thanks much.
 
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