3D Cartestian Coordinate Vector Problem

[FaraDazed]

EDIT: There was an issue where half of the post was missing, so I apologise but i have redone it

I was not sure if this was the most appropriate forum for this or not, so feel free to move if needed.

1. Homework Statement
A football (soccer ball) with a diametre of 0.44m is tracked using a motion tracking camera system and is found to have a trajectory of

$x=(-2.5+5t)m \\<br /> y=(2+3t)m \\<br /> z=(0.5 +2t-5t^2)m$

There are a few parts to the question and the part before the one I am stuck on asked to find the time when the centre of the ball touches the ground, so what I did was set z=0.22 and solved for t using quadratic formula which produced the answer 0.513 s which I am happy with. The next part asks:

What is the velocity immediately before it hits the ground?

Homework Equations

N/A[/B]

The Attempt at a Solution

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It is this bit which I am a bit stuck on, mainly because the way the question is set out is different to any similar kind of problems I have done previously. The only thing I can think of is differentiating each of those equations for the x, y and z components to get the x,y and z components of the velocity but that would only leave the z component as a function of time, I suppose that doesn't matter, but its the only thing I can think of, even though we have not covered this in the first two lectures we have had of this module.

Does this make sense?

$\frac{dx}{dt} = v_x = 5 \\<br /> \frac{dy}{dt} = v_y = 3 \\<br /> \frac{dz}{dt} = v_z = 2-10t \\$

And then @ t=0.513
$v_z = 2 - 10(0.513)=-3.13 \\<br /> \, \\<br /> \therefore \vec{v}= (5 \hat{i} + 3 \hat{j} - 3.13 \hat{k} ) m/s \\<br /> \therefore |\vec{v}|=\sqrt{5^2+3^2+3.13^2} = 6.62 m/s<br />$
Any help is really appreciated.

Thanks :)

 

[Orodruin]

You know the time it hits. What is the relationship between position and velocity? How does it look at an arbitrary time?

 

[FaraDazed]

You know the time it hits. What is the relationship between position and velocity? How does it look at an arbitrary time?

Yes I do know the time it hits (as long as I worked it out correctly, but for the sake of argument let's say that I did). Velocity is the derivative of postiion with respect to time?? What do you mean by how does it look at an arbitrary time?

EDIT: I just realized maybe you only say the post when half of it was missing? Something happened when I first posted it and half of the post got deleted, but its back there now. If not and you saw it as you see it now then I apologise.

 

[Orodruin]

Yes, I posted before your edit of the OP. Your approach looks reasonable. Without a force acting in the x and y directions, the velocity components in those directions will not change.

 

[FaraDazed]

Yes, I posted before your edit of the OP. Your approach looks reasonable. Without a force acting in the x and y directions, the velocity components in those directions will not change.

Ah right ok, thanks for taking a look :)