Question 1011969
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When the height (y)=0, the ball is at minimum or maximum distance
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{{{y=x-x^2/50}}}
Let y=0:
{{{0=x-x^2/50}}}
{{{0=50x-x^2}}}
{{{0=(x)(50-x)}}}
{{{x=0}}} {{{OR}}} {{{50-x=0}}}
{{{x=0}}} {{{OR}}} {{{50=x}}}
The minimum horizontal distance (start) is 0,
the maximum horizontal distance (ball hits ground) is 50.