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January 31, 2015

January 31, 2015

Posted by **Lola** on Friday, November 27, 2009 at 2:15am.

- Physics -
**MathMate**, Friday, November 27, 2009 at 3:12pm"a stone tied at the end of a 0.71 m long string"

r=0.71 m

"4 revolutions per second"

ω=4*2π radians/s

Tangential speed, v0

=rω

"The stone is 0.93 m above the ground"

Distance above ground, h

= 0.93

"an angle of 54 degrees above the horizontal"

θ=54°

Vertical component of initial velocity, vv

=v0sinθ

Let t=time it takes to hit ground,

S=vv*t+(1/2)gt²

-0.93 = vv*t+(1/2)(-9.8)t²

Solve for t.

I get -0.051 sec. or 4 sec. approx.

We reject -0.051 because t>0.

- Physics -
**Lilly**, Friday, November 27, 2009 at 4:57pmI tried doing the problem your way, but I got confused at the end.

For the equation S = w*t + (1/2)(-9.8)t**2, what was the value of w?

When I tried using w = v0 (17.844m/s), I got the same answers as yours

But when I tried using w = v0sin54 (14.4363m/s), I got different answers. I'm not sure which way is right. I'm leaning toward the latter one.

- Physics -
**MathMate**, Friday, November 27, 2009 at 5:44pmYou're quite right.

I have calculated for 5 revolutions per second. That is why they don't match your answers, which are on the right track.

I get now -0.06 and 3 seconds, and -0.06 is to be rejected.

What do you get for your answers?

Sorry for the inconvenience.

- Physics -
**Lilly**, Saturday, November 28, 2009 at 1:40amYes, I also got -0.06 and 3.00 seconds, so obviously the answer is 3.00 seconds.

Thanks for your help :)

- Physics -
**MathMate**, Saturday, November 28, 2009 at 8:21amYou're welcome!

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