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.

An example problem is shown below enter image description here enter image description here


2 Answers 2


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.

  • 1
    $\begingroup$ Thank you. The assumption when dividing through by t was what I did not understand/realise. $\endgroup$
    – K-Feldspar
    Commented Aug 11, 2016 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
so $t=0$ is one solution and $t=-a/b$ is the other.

  • $\begingroup$ 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 $y=0$. $\endgroup$
    – xish
    Commented Aug 11, 2016 at 21:11

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