How to calculate launch angle without initial velocity? I'm trying to calculate the launch angle of a projectile given:

*

*initial height,

*distance travelled,

*time of flight.

I feel like this should be possible, but I haven't been able to find any projectile formulas with only launch angle, height, distance and time as variables. Is this possible? If so, what would the formula look like? Thanks!
 A: If we assume your initial $x$ coordinate is $0$, the equations of motion will be:
\begin{equation}
y = y_0 + v \sin (\alpha) t - \frac{1}{2} g t^2
\end{equation}
and
\begin{equation}
x = v \cos (\alpha) t
\end{equation}
If you take the floor to be at height $y = 0$, the time it takes to reach the floor $T$ can be obtained from the first equation as $T(y_0, v, \alpha)$. Equivalenty, you can find $v (T, y_0, \alpha)$.
From the second equation, the distance traveled should be:
\begin{equation}
D = v \cos(\alpha) T
\end{equation}
So if I now put the velocity I found from the first equation in the last one, I'll have $D (T, \alpha, y_0)$, or equivalent, $\alpha(D, T, y_0)$.
So it's definitely possible. Try using this reasoning to derive the formula :).
A: This answer is a rephrasing of the previous answer to make it more accessible to the OP.
Given the initial height, $y_0$, horizontal distance, $D$, and time of flight, $T$, of a projectile, the vertical component of the velocity of the projectile, $v\sin\alpha$, may be calculated using the equation $0=y_0+v\sin\alpha \cdot T=\frac{1}{2}gT^2$, while the horizontal component of the velocity of the projectile, $v\cos\alpha$, may be computed using the equation $D=v\cos\alpha\cdot T$. The elementary trigonometric identity $\sin^2\theta+\cos^2\theta=1$ for all $\theta\in\mathbb{R}$ can then be employed to obtain the launch speed $v$ and the launch angle from the definition $\tan\alpha=\frac{v\sin\alpha}{v\cos\alpha}$.
