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anonymous
 one year ago
Physics 105:
Projectile problemfinding time
anonymous
 one year ago
Physics 105: Projectile problemfinding time

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to solve for time we don't really need to worry about the x component because just like when you throw an object in a moving can, it lands right back in your hand as if you were on a still ground and form your frame of reference it only moved in the 'y axes' now one more thing we need to do is know that the final y displacement is almost always going to be 0. I am also going to assume that you know the equation \[y _{f}=1/2g t^{2}+V _{o}Sin \theta+y _{o}\] So this equation should look like a quadratic equation to you where (a) is 1/2g, (b) is VoSintheta and (c) is Yo. you also need to know the quadratic formula because just like (x) time is our unknown here. the quadratic equation is; \[\frac{ b+/\sqrt{b ^{2}4ac} }{ 2a }\] substituting our values into this equation will give us; \[\frac{ V _{o}Sin \theta+/\sqrt{V _{o}^{2}Sin ^{2}\theta4(1/2g)(y _{o})} }{ 2(1/2g) }\] this will be equal to time.
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