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

A daredevil attempts to swing from the roof of a 24m high bulding to the bottom of an identical building using a 24m rope. He starts from rest with the rope horizontal, but the rope will break if the tension force in it is twice the weight of the daredevil. How high is the rope above level when it breaks? (Hint: Apply the law of conservarion of energy) Also, the building are 24m apart.

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

  • Physics - please help I don't get it - , Sunday, March 1, 2009 at 3:31pm

    please I really need help with this

  • Physics - , Sunday, March 1, 2009 at 4:21pm

    The 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 - , Sunday, March 1, 2009 at 4:41pm

    wow, that was more elaborate than I thought it would be. Thank you so much for answering my qustion. :]

  • Last lines irrelavant - , Sunday, March 1, 2009 at 4:50pm

    These were just to keep my geometry in mind.
    cos A = 1
    A = 0
    That is right at the ground.

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