A package is released from an aircraft flying level. Its time of fall is found from:

1. Released from level flight, the package starts with zero vertical velocity, so $\theta=0$. 2. The vertical equation $y=v_0\sin\theta\,t+\tfrac12 gt^2$ therefore reduces to $y=\tfrac12 gt^2$. 3. Rearranging for $t$ gives $t=\sqrt{\dfrac{2y}{g}}$. 4. The formula $t=\dfrac{v_0\sin\theta}{g}$ applies to a projectile launched at an angle, and here it evaluates to zero. 5. Missing that a horizontal projection has $\theta=0$ was the recorded error on this question. _Source: NECTA ACSEE 2023 Physics 131/1, Question 2: Mechanics (Projectile Motion)_