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March 3, 2015

March 3, 2015

Posted by **Aaron** on Sunday, March 1, 2009 at 2:49pm.

I don't understand; how can I do this without knowing the daredevil's weight?

- Physics - please help I don't get it -
**Aaron**, Sunday, March 1, 2009 at 3:31pmplease I really need help with this

- Physics -
**Damon**, Sunday, March 1, 2009 at 4:21pmThe falling distance as a function of angle is f = 24 cos A

where A is the angle between the line and the wall of the target building. It starts out at A = 90 and if he got there, ends at A = 0

The component of the weight along the line is m g cos A

(1/2) m v^2 = m g f = m g (24 cos A)

so

v^2 = 48 g cos A

The centripetal acceleration is

v^2/r = 2 g cos A

The tension in the line at breaking = 2 m g + m g cos A which is the mass times the centripetal acceleration

T - m g cos A = 2 m g cos A

T = 3 m g cos A

so when T = 2 m g

2 m g = 3 m g cos A

cos A = 2/3

A = 48.2 deg

distance fallen = 24 cos 48.8

= 24(2/3) = 16

an 24 - 16 = 8 above ground

cos A = 1

A = 0

That is right at the ground.

- Physics -
**Aaron**, Sunday, March 1, 2009 at 4:41pmwow, that was more elaborate than I thought it would be. Thank you so much for answering my qustion. :]

- Last lines irrelavant -
**Damon**, Sunday, March 1, 2009 at 4:50pmThese were just to keep my geometry in mind.

cos A = 1

A = 0

That is right at the ground.

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