We are told that the ball is thrown upwards, so let the upwards direction be positive and gravity will act in the opposite direction, thus $a=-9.8$. At the ball's greatest height, $s=\\var{s}$, the velocity will be $v=0$ for an instant as the ball changes direction and falls downwards. We are trying to find the initial velocity of the ball, $u$. We have everything except $t$ so we can use the formula $v^2 = u^2 + 2as$, rearranged for $u$.
\n\\begin{align} v^2 & = u^2 + 2as, \\\\
0 & = u^2 + \\left(2 \\times \\var{a} \\times \\var{s}\\right), \\\\
0 & = u^2 - \\var{2*-a*s}.
\\end{align}
Therefore
\n\\begin{align} u^2 &= \\var{2*-a*s}, \\\\
u & = \\var{precround ((2*-a*s)^(1/2),3)} \\mathrm{ms^{-1}}.
\\end{align}
The initial velocity of the ball is $ \\var{precround ((2*-a*s)^(1/2),3)} \\mathrm{ms^{-1}}.$
\nWe have now that $s=\\var{distance}$, $u = \\var{u}$ and $a=\\var{a}$. To find the time the ball is above $\\var{distance} \\mathrm{m}$ we need to find the time the ball first passes over $\\var{distance} \\mathrm{m}$ and the time it then falls back down past $\\var{distance} \\mathrm{m}$, and find the difference between those two times. To find these times we can use the formula $s=ut+\\frac{1}{2}at^2$, rearranged for $t$.
\nWe have
\n\\begin{align} s & = ut + \\frac{1}{2}at^2, \\\\
\\var{distance} & = \\var{u}t - \\left( \\frac{1}{2} \\times \\var{-a} \\times t^2 \\right), \\\\
4.9t^2 - \\var{u}t + \\var{distance} & = 0.
\\end{align}
Therefore our first $t$ value, when the ball first passes $\\var{distance} \\mathrm{m}$, is
\n\\begin{align} t & = \\frac{\\var{u} + \\sqrt{\\var{u}^2 - \\left(4 \\times 4.9 \\times \\var{distance} \\right)} }{2 \\times 4.9}, \\\\
& = \\var{t1} \\mathrm{s},
\\end{align}
and our second $t$ value, when the ball second passes $\\var{distance} \\mathrm{m}$, is
\n\\begin{align} t & = \\frac{\\var{u} - \\sqrt{\\var{u}^2 - \\left(4 \\times 4.9 \\times \\var{distance} \\right)} }{2 \\times 4.9}, \\\\
& = \\var{t2} \\mathrm{s}.
\\end{align}
Then the total time for which the ball is $\\var{distance} \\mathrm{m}$ or more above $X$ is $\\var{t1} - \\var{t2} = \\var{t1-t2}$ seconds.
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