I'm really new to all this, so please don't be rude. I have here what appears to be a simple formula. Yet I cant get my head around it:

$$\begin{align} y &= y_0 + v_{0y}t + (1/2)a_y t^2 \\ 0 &= 1.5 + 76.95t + (1/2)(-9.8)t^2 \end{align}$$

I am confused as to how to: SOLVE $t$.

I am confused, because I am not sure how to calculate this, because there are almost no operators present: like + - * / between the brackets or the t.

Could you please provide me with a detailed response in regards to solving this issue.


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  • 2
    $\begingroup$ Are you familiar with the method of solving quadratic equations? $\endgroup$ – Kwin van der Veen Jan 7 '14 at 0:45
  • $\begingroup$ I understand simple problems, but this one is just too much $\endgroup$ – TastyLemons Jan 7 '14 at 0:47
  • 1
    $\begingroup$ Those missing operators are multiplications... $\endgroup$ – DJohnM Jan 7 '14 at 0:53

If you want to solve for the time when $y=0$, you have the following equation $$ \frac{a_y}{2}t^2 + v_{0y}t + y_0 = 0 $$

Let's forget the numbers for now and work with variables only. This is a quadratic equation; you can't solve it like linear equations you might be familiar with. For this, you need a formula, given here or here. Using those formulas, the solutions for $t$ are $$ t = \frac{-v_{0y}\pm\sqrt{v_{0y}^2-2a_y y_0}}{a_y} $$

Now plug in numbers and find $t$. Note that there are two solutions, one each associated with the sign ($\pm$). One represents when the projectile hits the ground in the future, and the other a hypothetical time in the past when the projectile was at ground level if you traced the trajectory backwards.


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