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April 20, 2014

April 20, 2014

Posted by **jeff** on Monday, December 22, 2008 at 12:55am.

Can you check my work

Mgh - Mg(2R) = 0.5MV^2

gh - 2gR = 0.5V^2

gh-2gR = 0.5Rg

h = 2.5R = 2.5(.36m)= 0.9

for the second part to figure out how far from the table it lands:

Y = Voyt + .5gt^2

0.9 = 4.9t^2

t = 0.428571s

x = Voxt

x =((Rg)^0.5)*t

- physics -
**Damon**, Monday, December 22, 2008 at 6:34amYour method is correct. I did not check the arithmetic but do not see any obvious errors.

- physics -
**drwls**, Monday, December 22, 2008 at 6:47amIf it just makes the loop without losing contact, V^2/R = g at the top of the loop (y = 2R). The highest elevation must be such that

m g (h-2R) = (1/2)m V^2 = (1/2)m g R

h = (5/2) R is correct

I don't understand what you did after that.

At the bottom of the loop, I assume that the velocity (V') starts out horizontal and satisfies the equation

(1/2)mV'^2 = 2.5 m R g

V' = sqrt (5Rg) = 4.2 m/s = Vox

The time t that it spends falling satisfies

(1/2) g t^2 = 0.9 m

t = sqrt(1.8/9.8) = 0.43 s

Vox*t = 1.8 m

It is possible that the object is "launched" at some angle other than horizontal. That would depend upon how the loop was constructed. No sketch was provided.

- physics -
**drwls**, Monday, December 22, 2008 at 6:51amIn your second part, I agree with your t calculation but not your Vox

- physics -
**drwls**, Monday, December 22, 2008 at 6:51amIn your second part, I agree with your t calculation but not your Vox

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