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April 18, 2015

April 18, 2015

Posted by **Sam** on Thursday, September 23, 2010 at 2:34pm.

- Physics -
**bobpursley**, Thursday, September 23, 2010 at 2:44pmWell, you have some algebra ahead.

Vertical:

155=VsinTheta*t-4.9t^2

Horizontal:

195=VcosTheta*t

you are given t.

So you have two equations, two unknowns.

Here is what I suggest. Put both equations in the form of

V*trigfunction=some number

then square both sides.

add the equations, you should get something like this:

V^2(cosTheta ^2 + sinTheta ^2)=number^2

well, you know the ( ) is 1, which allows you to solve for V. Then substitute back to solve for theta.

Have fun.

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