The horizontal distance x and the vertical height y of a projectile at a time t are given by

                     $x=at and y= b{t}^2+ct$

where a, b, and c are constants. The magnitude of the initial velocity of the projectile is given by

The horizontal and vertical components of velocity are

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{at}\right)=\mathrm{a} \left(\mathrm{i}\right) \\ {\mathrm{v}}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}({\mathrm{bt}}^2+\mathrm{ct})=2\mathrm{bt}+\mathrm{c} \left(\mathrm{ii}\right) \end{gathered}$$

If a projectile is projected with an initial velocity u at an angle θ with the horizontal, the horizontal and vertical components of its velocity at time t are given by

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}=\mathrm{u} \cos \mathrm{\theta } \left(\mathrm{iii}\right) \\ \mathrm{and} {\mathrm{v}}_{\mathrm{y}}=\mathrm{u} \sin \mathrm{\theta }-\mathrm{gt} \left(\mathrm{iv}\right) \end{gathered}$$

Comparing (iii) and (iv) with (i) and (ii) above we have;

$$\begin{gathered} u \cos \theta =a and u \sin \theta =c. \\ Squaring and adding, we get: \\ u^2=a^2+c^2 \\ or u={(a^2+c^2)}^{1/2} \end{gathered}$$