• anonymous
a block is shot up 32 degrees friction less incline with initial velocity of 3.5 m/s How far does it go
  • Stacey Warren - Expert
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  • katieb
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  • anonymous
|dw:1447140129672:dw| Here is the situation we have. The first thing we should notice is that we can apply the conservation of energy. We initially have only kinetic and end up with some potential energy, thus\[\huge \frac{ 1 }{ 2 }mv^2 = mgh \implies \frac{ 1 }{ 2 }v^2=gh\] We see that the mass cancels out on both sides! Here comes the tricky part! In this equation, we have h, which is the HEIGHT at which it is vertically displaced. It doesn't give us the distance S (labeled in diagram) that it travels. Well this is a bummer! But no worries, we can use trig to re-express the height. As you can see, the height is the leg that is opposite of the 32deg angle! Therefore, we can say, using basic geometry principles, that the hypotenuse times the sin of the angle is equal to the height\[\huge h=S\sin(32^o)\] Substituting this into our equation, we can see that \[\huge \frac{ 1 }{ 2 }v^2=gS\sin(32^o)\] You know v, and you know g, now solve for S X)
  • anonymous
@Michele_Laino I keep overthinking it... we can do this, yea?
  • Michele_Laino
correct! @CShrix

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