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a)

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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.

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We 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.

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Therefore we can use the equation $v^2=u^2+2as$, rearranged for $s$.

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Note that the rooftop is $\\var{s} \\mathrm{m}$ above ground level, so the distance travelled by the ball to its greatest height is $(\\simplify{s-{s}}) \\mathrm{m}$.

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\\begin{align} v^2 & =u^2+2a(s-\\var{s}), \\\\
                     2a(s-\\var{s}) & = v^2-u^2, \\\\
                         s-\\var{s} & = \\frac{v^2-u^2}{2a}, \\\\
                           & = \\frac{0 - \\var{u}^2}{2 \\times \\var{a}}, \\\\
                           & = \\var{precround(-u^2/(2*a),3)}, \\\\
s &= \\simplify[]{{s}+{precround(-u^2/(2*a),3)}}, \\\\
&= \\var{precround(max_height,3)} \\mathrm{m}.
\\end{align}

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The greatest height reached by the ball is $\\var{precround(max_height,3)} \\mathrm{m}$.

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b)

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The flight time of the ball is the length of time between it being thrown from the rooftop and when it hits the ground below.

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Here, the rooftop is $\\var{s}\\mathrm{m}$ above the ground, therefore the ball will stop moving when it hits the ground at a height $\\var{-s} \\mathrm{m}$ from its original position.

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Therefore we can use the equation $s=ut+\\frac{1}{2}at^2$ rearranged for $t$.

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\\begin{align} s & = ut + \\frac{1}{2}at^2, \\\\
                    -\\var{s} & = \\var{u}t - \\frac{49}{10}t^2, \\\\
                        \\frac{49}{10}t^2 - \\var{u}t - \\var{s} &= 0. 
\\end{align}

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Using the quadratic formula, taking only the positive term as we are modelling time, this gives 

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\\begin{align} t & = \\frac{u + \\sqrt{u^2 +19.6s)}}{9.8}, \\\\
                       & = \\frac{\\var{u} + \\sqrt{\\var{u^2} + \\left(19.6 \\times \\var{s}\\right)}}{9.8}, \\\\
                       & = \\var{precround((u+ (u^2 + 19.6*s)^(1/2))/9.8,3)} \\mathrm{s}.
\\end{align}

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The flight time of the ball is $\\var{precround((u+ (u^2 + 19.6*s)^(1/2))/9.8,3)}$ seconds.                     

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Find the greatest height in $\\mathrm{m}$ (to 3d.p.) above the ground reached by the ball.

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You have not given your answer to the correct precision.

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The time in $mathrm{s}$ (to 3d.p.) of flight of the ball, before it lands on the ground.

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You have not given your answer to the correct precision.

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A ball is thrown vertically upwards with a speed $\\var{u} \\mathrm{ms^{-1}}$ from a rooftop which is $\\var{s} \\mathrm{m}$ above ground. The acceleration due to gravity is $g = 9.8 \\mathrm{ms^{-2}}$.

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Acceleration due to gravity with upwards as positive direction

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time to 3d.p.

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distance of rooftop above ground

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initial velocity ball is thrown at

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A ball is thrown vertically upwards from a position above ground level. Find its greatest height, and the total time its in the air.

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