## anonymous one year ago Physics 105: Projectile problem-finding time

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.