A cannonball is fired from a cannon with a velocity of $24.0\,\frac{\mathrm m}{\mathrm s}$ at an angle of $40^\circ$ relative to the positive $x$-axis. It lands on the ground $56.6\,\mathrm m$ from where it was launched. How much time did the cannonball spend in the air? (Ignore air resistance.)
I can solve for the airtime $t$ by using either position equation:
$$\begin{cases} x=v_0\cos40^\circ t\\[1ex] y=v_0\sin40^\circ t-\frac g2t^2 \end{cases}$$
Applying the first equation, I get
$$56.6\,\mathrm m=\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t\approx3.08\,\mathrm s$$
On the other hand, applying the second equation, I get
$$\begin{align*} 0&=\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ t-\frac g2t^2\\[1ex] \implies0&=\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ-\frac g2t\\[1ex] \implies t&=\frac{2\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ}g\approx3.15\,\mathrm s \end{align*}$$
assuming $g=9.80\,\frac{\mathrm m}{\mathrm s^2}$. These differing times suggest to me that the cannon reached a distance along the ground of $56.6\,\mathrm m$ before it hits the ground. What is the source of the discrepancy?
My initial thought is that it has to do with what value of $g$ we agree to use. If I were to set $t=3.08\,\mathrm s$ as found in the $x$ equation and substitute this into the $y$ equation to solve for an unknown $g$, I find that the "correct" gravitational acceleration is closer to $10.0175\,\frac{\mathrm m}{\mathrm s^2}$.
Alternatively, I can consider the cannonball's vertical velocity $v_y=v_{0y}-gt$ to determine the time it takes to reach its peak height, at which point $v_y=0$:
$$0=\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ-gt\implies t\approx1.57417\,\mathrm s$$
Then it should take the cannonball twice this time to reach the ground, which suggests that it would stay in the air for about $3.15\,\mathrm s$. Is the result of any one of these equations more accurate?
Perhaps I should just take the average of the two times I found?