A man throws a ball to maximum horizontal distance of $80\,m$. Calculate the maximum height reached.

Hint The height of the ball can be determined by using two formula of the projectile motion. First by using the range of the projectile motion formula the velocity is determined. Then using this velocity value in the height of the projectile motion formula, the height of the ball can be determined.

Useful formula
The range of the projectile motion is given by,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Where, $R$ is the range of the ball, $u$ is the velocity of the ball, $\theta $ is the angle for the maximum height and $g$ is the acceleration due to gravity.
The height of the projectile motion is given by,
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Where, $H$ is the height of the ball, $u$ is the velocity of the ball, $\theta $ is the angle for the maximum height and $g$ is the acceleration due to gravity.

Complete step by step solution
Given that,
The maximum horizontal distance or range is, $R = 80\,m$
Now,
The range of the projectile motion is given by,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}\,................\left( 1 \right)$
By substituting the range value in the above equation (1), then the above equation (1) is written as,
$80 = \dfrac{{{u^2}\sin 2\theta }}{g}$
For the maximum height, the ball should be thrown in the angle of ${45^ \circ }$, then the above equation is written as,
$80 = \dfrac{{{u^2}\sin 2 \times {{45}^ \circ }}}{g}$
By multiplying the terms, then
$80 = \dfrac{{{u^2}\sin {{90}^ \circ }}}{g}$
From trigonometry the value of the $\sin {90^ \circ } = 1$, then
$80 = \dfrac{{{u^2}}}{g}\,.................\left( 2 \right)$
Now,
The height of the projectile motion is given by,
$H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}\,......................\left( 3 \right)$
By rearranging the terms in the above equation, then
$H = \dfrac{{{u^2}}}{g}\dfrac{{{{\sin }^2}\theta }}{2}$
By substituting the equation (2) in the above equation, then
$H = 80 \times \dfrac{{{{\sin }^2}\theta }}{2}$
By substituting the angle of ball thrown, then
$H = 80 \times \dfrac{{{{\left( {\sin {{45}^ \circ }} \right)}^2}}}{2}$
In trigonometry ${\sin ^2}\theta = {\left( {\sin \theta } \right)^2}$,
The value of $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, then
$H = 80 \times \dfrac{{{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}}}{2}$
By squaring the terms, then
$H = 80 \times \dfrac{{\left( {\dfrac{1}{2}} \right)}}{2}$
By rearranging the terms, then
$H = \dfrac{{80}}{{2 \times 2}}$
On dividing the terms, then
$H = 20\,m$

The maximum height reached by the ball is $20\,m$.

Note: For every object, if we want to throw the object which goes to the maximum distance for the given velocity, the angle to throw the object in ${45^ \circ }$, as we release the object in ${45^ \circ }$, the object will definitely reach the maximum distance, for the given velocity.