# Projectile: know only launch velocity, max height, and distance

Suppose you know only these three things about a launched projectile:

- Initial launch velocity (magnitude only, not direction)
- Maximum height reached
- Horizontal distance traveled before hitting the ground

Is it possible to find the initial height, launch angle, and airtime of this projectile (assuming that the initial launch height may or may not be at ground level)?

Yes it is possible. You could use a variation of the method I suggested in a similar thread you started.

I did what you said, and the good news is that I came up with the following equation:

$\frac{d^{2}g}{4H - 2h + 4\sqrt{H(H-h)}} = v^{2} - 2gH + 2gh$

The bad news is that I need to solve for the initial height, $h$ in terms of all the other variables, and there seems to be no easy way to make things work. I tried assigning the term $r = H-h$, but it's still a huge pain to solve

If you are willing to assume initial launch height is zero or negligibly small...

θ=(1/2)arcsin(g*x_max/v^2)
=arcsin[sqrt(2*g*y_max)/v]

t_flight=2*sqrt[2*y_max/g]

Otherwise it's ugly. I see mathematics as both an enabler and a disabler, in your problem it is definitely the latter :(

The equation I get is $$\frac {a} {\sqrt {b - z}} - \sqrt {z} - 1 = 0$$ where $$a = \frac {d} {2H}$$$$b = \frac {v^2} {2gH}$$$$z = 1 - \frac h H$$ This can be solved numerically. Or, after some more massage, it could be converted into a quartic equation that could be solved by Ferrari's method, or numerically. The latter is bit tricky, because the quartic has four roots, but there is only one physical solution. The physical solution has these properties: $z \ge 0$ (because $h \le H$) and $z \lt b$ (because initial kinetic energy must be greater than required for purely vertical motion from $h$ to $H$). Note also that not all combinations of parameters admit a solution.

What are d, H, and h? I don't see where you said this.

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Distance traveled, max height, launch height.