# 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!

• What does height refer to here ? Is it the maximum height that the projectile achieves ? Commented Apr 23, 2021 at 15:09
• 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? Commented Apr 23, 2021 at 15:15
• Does the projectile land at the same elevation that it was launched from? Commented Apr 23, 2021 at 21:15

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

• This is exactly what I was looking for. Thanks for the thorough description! Commented Apr 26, 2021 at 14:51

If we assume your initial $$x$$ coordinate is $$0$$, the equations of motion will be: $$$$y = y_0 + v \sin (\alpha) t - \frac{1}{2} g t^2$$$$

and $$$$x = v \cos (\alpha) t$$$$

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:

$$$$D = v \cos(\alpha) T$$$$

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 :).