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    serenitymantra's Avatar
    serenitymantra Posts: 28, Reputation: 1
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    #1

    Sep 22, 2008, 03:31 PM
    Velocity at 30 degree angle
    I want to check if I am on the right track.

    Question: Santa loses his footing and slides down a frictionless snowy roof that is tilted at 30 degrees. If Santa slides 10m before reaching the edge, what is his speed as he leaves the roof?

    Would I use ax=g sin (theta) and then take my results and plug it in (Vx)f= square root of ax (delta x)?

    Thanks!
    Credendovidis's Avatar
    Credendovidis Posts: 1,593, Reputation: 66
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    #2

    Sep 22, 2008, 04:16 PM
    Hello SM ,

    Why difficult if you can do it easy ?

    Question: Santa loses his footing and slides down a frictionless snowy roof that is tilted at 30 degrees. If Santa slides 10m before reaching the edge, what is his speed as he leaves the roof?

    See the roof as the 10 m long hypothenusa of a 30 degree triangle.
    ( 1, 2, V3 )
    The vertical side of that triangle is 10/2 meter = 5 meter.
    So all what remains is to calculate the speed after a fall of 5 meter (as the roof is frictionless) at 1 G (10 meter/sec).

    I leave you to that yourself as it is YOUR homework!!

    ;)
    serenitymantra's Avatar
    serenitymantra Posts: 28, Reputation: 1
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    #3

    Sep 22, 2008, 04:44 PM
    Quote Originally Posted by Credendovidis View Post
    Hello SM ,

    Why difficult if you can do it easy ?

    Question: Santa loses his footing and slides down a frictionless snowy roof that is tilted at 30 degrees. If Santa slides 10m before reaching the edge, what is his speed as he leaves the roof?

    See the roof as the 10 m long hypothenusa of a 30 degree triangle.
    ( 1, 2, V3 )
    The vertical side of that triangle is 10/2 meter = 5 meter.
    So all what remains is to calculate the speed after a fall of 5 meter (as the roof is frictionless) at 1 G (10 meter/sec).

    I leave you to that yourself as it is YOUR homework !!!

    ;)

    Hi, I know. I don't expect answers, just a push in the right direction. I really want to learn this stuff. :) It's just been tough because our professor moves really fast & it's his first time teaching.

    What I didn't follow is why it's 10/2 for the veritcal angle.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Sep 23, 2008, 05:36 AM
    Quote Originally Posted by serenitymantra View Post

    What I didn't follow is why it's 10/2 for the veritcal angle.
    sin(30 degrees) = 1/2. So the vertical drop along the 10 meters of roof is 1/2* 10 = 5 meters.

    You can do this just like the other one that you posted - Santa's acceeration is 9.8m/s * sin(30). Then use the formula v^2 = 2ad to solve for v.
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