Refer to the diagram shown below.
The following discussion assumes a simplistic analysis that ignores air resistance and variations in the terrain that the missile travels over.
Let the launch velocity be V₀ at an angle of θ relative to the horizontal.
The horizontal component of velocity is V₀ cosθ.
If the time of flight is [tex]t_{f}[/tex], then
[tex]r=V_{o} \, t_{f}[/tex]
where r = the range of the missile.
Also, the time, t, when the missile is at ground level is given by
[tex]0=V_{o} sin\theta \, t- \frac{1}{2}gt^{2} [/tex]
where g = acceleration due to gravity.
t = 0 corresponds to when the missile is launched. Therefore
[tex]t_{f} = \frac{2V_{o}sin\theta}{g} [/tex]
Therefore
[tex]r= \frac{2V_{o}^{2} sin\theta cos\theta}{g} = \frac{V_{o}^{2} sin(2\theta)}{g} [/tex]
Typically, θ=45° to achieve maximum range, so that
[tex]r= \frac{V_{o}^{2}}{g} [/tex]
This analysis is more applicable to a scud missile rather than a powered, guided missile.
Answer:
[tex]t_{f} = \frac{r}{V_{o} cos\theta} \\\\ r= \frac{V_{o}^{2} sin(2\theta)}{g} [/tex]
Usually, θ=45°