You can model the ball as a particle moving in a straight line with constant acceleration of magnitude $9.8 \\mathrm{ms}^{-2}$ due to gravity. As the ball is thrown upwards, we take the positive direction as pointing upwards.
\nWe are told $u=\\var{u}$ and $a=\\var{a}$ and we are asked to find the greatest height, $s$, reached by the ball. At its greatest height, the ball will for an instant have a velocity of $v=0$ before it starts to fall back down.
\nTherefore we can use the equation $v^2=u^2+2as$, rearranged for $s$.
\n\\begin{align} v^2 &= u^2+2as\\\\
2as &= v^2-u^2\\\\
s &= \\frac{v^2-u^2}{2a}\\\\
&= \\frac{0 - \\var{u}^2}{2 \\times \\var{a}} = \\var{max_height}\\\\
\\end{align}
The greatest height reached by the ball is $\\var{precround(max_height,3)}$ metres
\nWe can use the same equation to find the speed of the ball when it reaches a given height
\n\\begin{align} v^2 &= u^2+2as\\\\
&= (\\var{u})^2+(2 \\times \\var{a} \\times \\var{s})\\\\
&= \\var{v_ss}\\\\
\\end{align}
The speed of the ball when it reaches a height of $\\var{s}$ metres is $\\var{precround(v_s,3)}\\,ms^{-1}$
\nc)
\nThe total time that the ball is above a given height $(s)$ can be found in a number of ways. The best method is to use the equation $s=ut+\\frac{1}{2}at^2$. This is a quadratic equation in $t$ and will have two solutions for $t$ - one will be the time on the way up and the other, the time on the way down. Subtracting the \"up\" time from the \"down\" time will give the total time above that height.
\n\\begin{align} s & = ut + \\frac{1}{2}at^2\\\\
\\frac{1}{2}at^2 + ut - s & = 0\\\\
\\end{align}
Using the quadratic formula gives
\n\\begin{align} t &= \\frac{-u \\pm \\sqrt{u^2 +2as}}{a}\\\\
&= \\frac{\\var{-u} \\pm \\sqrt{\\var{u^2} + \\left(2 \\times \\var{a} \\times \\var{d}\\right)}}{\\var{a}}\\\\
t &= \\var{precround(t1,3)}\\,\\,\\,\\,\\mathrm{or}\\,\\,\\,\\,\\var{precround(t2,3)}\\\\
\\end{align}
The total time that the ball is at, or above, $\\var{d}$ metres is $\\var{precround(t,3)}$ seconds
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\n\nModelling the ball as a particle and assuming air resistance is negligible, find
\n(a) the maximum height reached by the ball
\n(b) the speed of the ball when it is $\\var{s}\\,$ metres above ground
\n(c) the total time the ball is at least $\\var{d}\\,$ metres above the ground
\n\nThe acceleration due to gravity is $g = 9.8 \\mathrm{ms^{-2}}$
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