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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!

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  • $\begingroup$ What does height refer to here ? Is it the maximum height that the projectile achieves ? $\endgroup$ Commented Apr 23, 2021 at 15:09
  • $\begingroup$ There doesn't always have to exist a single equation for every situation, have you considered using several of the kinematics equations at the same time? $\endgroup$
    – Triatticus
    Commented Apr 23, 2021 at 15:15
  • $\begingroup$ Does the projectile land at the same elevation that it was launched from? $\endgroup$ Commented Apr 23, 2021 at 21:15

2 Answers 2

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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 $\tan\alpha=\frac{v\sin\alpha}{v\cos\alpha}$.

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    $\begingroup$ This is exactly what I was looking for. Thanks for the thorough description! $\endgroup$ Commented Apr 26, 2021 at 14:51
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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 :).

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