Question

solve for t

Answer 1

\[y=v_0*t+\frac{ g*t^2 }{ 2 }, v_0\neq0\]

Answer 2

If \(y\), \(v_0\), and \(g\) are constants, then you can use the quadratic formula.

Answer 3

assuming that they aren't

Answer 4

because v isn't a constant, and neither y. g can be constant, but that can easily change to a as an acceleration

Answer 5

To my knowledge, \(v_0\) is initial velocity, so it's constant in a particular context, and \(g\) must be referring to the acceleration due to gravity, which is (for our purposes) constant.

But if \(y\) is a variable? Hmm, maybe you can try completing the square...

Answer 6

you are correct. and yes, y is a variable. I did complete the square, but what I got is not the same wolfram alpha is displaying…

Answer 7

\[\begin{align*}y&=v_0t+\frac{1}{2}gt^2\\
&=\frac{1}{2}g\left(t^2+\frac{2v_0}{g}t\right)\\
&=\frac{1}{2}g\left(t^2+\frac{2v_0}{g}t+\frac{v_0^2}{g^2}-\frac{v_0^2}{g^2}\right)\\
&=\frac{1}{2}g\left(\left(t+\frac{v_0}{g}\right)^2-\frac{v_0^2}{g^2}\right)\\
y&=\frac{1}{2}g\left(t+\frac{v_0}{g}\right)^2-\frac{v_0^2}{2g}
\end{align*}\]
Should be easy to isolate \(t\).

Answer 8

what I got is \[t=\sqrt{\frac{ 2y }{ g }}\]

Answer 9

that's a neat trick you did.

Answer 10

You should end up getting
\[t=\frac{-v_0\pm\sqrt{2yg+v_0^2}}{g}\]