Physics 105: Projectile problem-finding time

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Physics 105: Projectile problem-finding time

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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to 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; \[\frac{ V _{o}Sin \theta+/-\sqrt{V _{o}^{2}Sin ^{2}\theta-4(1/2g)(y _{o})} }{ 2(1/2g) }\] this will be equal to time.

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