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prosteve037

Joined: 25 Sep 2010
Posts: 6
Location: 222

 Posted: Fri Oct 15, 2010 2:51 am    Post subject: How Do I Get This Equation? Using vy = voy - gt and voy = vosinθ, I can get vy = vosinθ - gt and t = (vy - vosinθ) / -g Eliminating time, t, from the equation y = voyt - .5gt^2, how can I show that vy^2 = vo^2sin^2θ - 2gy? I just can't seem to figure out the algebra here :/ Please help
hepcj

Joined: 23 Jun 2007
Posts: 125

 Posted: Fri Oct 15, 2010 9:43 am    Post subject: Can you please post the question as written so I can see what information is given. However, to get the sin (2θ) you would need a cos(θ) somewhere, probably from the x-component of velocity since: sin (2θ) = sin(θ + θ) = cos(θ) sin(θ) + cos(θ) sin(θ) sin (2θ) = 2 cos(θ) sin(θ)
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