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So I have two similar questions:

$x=x_0 + v_0t + \frac{1}{2}at^2$ where I have to solve for $t$

and

$mgx + \frac{1}{2}kx^2 = \frac{1}{2}mv^2$ where I have to solve for $x$

I'm not sure if I have to use the quadratic equation or what.

anomaly
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1 Answers1

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Hint: First one:

$$\dfrac{1}{2}at^2 + v_0t + (x_0 - x) = 0$$ and apply quadratic equation to get $$t = \dfrac{-v_0 \pm \sqrt{v_0^2-4\left(\dfrac{1}{2}\right)a(x_0 - x)}}{2\left(\dfrac{1}{2}\right)a}$$

Now for the second one: $$\dfrac{1}{2}kx^2 + mgx - \dfrac{1}{2}mv^2 = 0$$

Same process as the first. Do note that $- \dfrac{1}{2}mv^2$ has the negative sign.

Clarinetist
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