Finding how long a projectile is in the air: why does y=0 give the time that it lands and not the time that it is launched?

When using the formula below to find the time the projectile is in the air, why does y (vertical displacement) = 0 return the time at which the ball lands, and not the time at which the ball is released? Both occur at $y=0$.

$$y=v_0\sin(\alpha)t+\frac12 g\,t^2$$

• Where $v_0$ is the initial velocity, $\alpha$ is the angle of release from horizontal, $g$ is the local gravitational acceleration and $t$ is time.

It does both. $t=0$ is a solution to that equation just as well:

$$y=v_0\sin(\alpha) t+\frac{1}{2}gt^2\quad \Leftrightarrow \quad 0=v_0\sin(\alpha) \cdot0+\frac{1}{2}g\cdot 0^2\Leftrightarrow0=0$$

The reason that they don't find that in the solution you are showing is, that they devide through with $t$ during their reduction. To do this, they silently assume $t\neq 0$.

Thereby the solution(s) they get cover all cases except the $t=0$ case. To complete it, this case therefore ought to be checked seperately. And by doing that (by inserting $t=0$), you'll find that $t=0$ is indeed another solution, which gives you two solutions in total.

This could be avoided by not doing the divide-through-with-$t$ step and instead just using the usual solution formula for a quadratic equation.

• Thank you. The assumption when dividing through by t was what I did not understand/realise. – K-Feldspar Aug 11 '16 at 21:10

The equation of the trajectory gives both times. It is a quadratic so it has 2 solutions for time $t$. It can be factorised into the form
$y=(a+bt)t$
so $t=0$ is one solution and $t=-a/b$ is the other.

• Just to add, the reason that one of the solutions of this equation is $t=0$ is that it is derived with the assumption that the projectile is launched at $t=0$ from $y=0$. – xish Aug 11 '16 at 21:11