91Ó°ÊÓ

The projectile's horizontal range is the maximum distance it travels horizontally before striking the ground, and it is proportional to the angle of projection.

The range of the projectile is given by,

$R_{max}=\frac{v_0^2\sin 2\mathrm{Î\pm }}{g}$

Here,$v_0$is the initial velocity,$\mathrm{Î\pm }$is the angle, g is the acceleration due to gravity.

Maximum height is given by,

$H=\frac{v_0^2{\sin }^2\mathrm{Î\pm }}{2g}$

The maximum horizontal range for angle of projection of 45.

$$\begin{gathered} R_{max}=\frac{v_0^2\sin 2\left(45°\right)}{g} \\ =\frac{v_0^2}{g} \end{gathered}$$ …(¾±)

The maximum vertical height angle 45 is given by,

$$\begin{gathered} H=\frac{v_0^2{\sin }^{2}45°}{2g} \\ =\frac{v_0^2}{4g} \end{gathered}$$ …(¾±¾±)

The relation between height and range is given by equation (i) and (ii),

$$\begin{gathered} H_{max}=\frac{R_{max}}{4} \\ 4H_{max}=R_{max} \end{gathered}$$

Thus, the relation of maximum horizontal range and maximum vertical height is $4H_{max}=R_{max}$.