a) We resolve $F=ma$ with vertically upwards being the positive direction, as the ball is projected upwards. Both the resistance and the force due to gravity are acting in the downwards direction, against the path of the ball.
\n\\begin{align} F & = ma \\\\
- \\var{resistance} - \\left(\\var{mass} \\times 9.8\\right) & = \\var{mass} \\times a \\\\
a & = \\frac{ - \\var{resistance} - \\left(\\var{mass} \\times 9.8\\right) } {\\var{mass}} \\\\
& = \\var{a} \\end{align}
The acceleration of the ball is $\\var{a} \\ \\mathrm{ms^{-2}}$.
\nb) When the ball reaches its greatest height it stops moving for an instant before it changes direction therefore $v=0$. Now we have $u = \\var{u}, a = \\var{a}$ and $v=0$ and we want to find $s$. We can use the formula $v^2 = u^2 + 2as$, 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{s}\\end{align}
The greatest height reached by the ball is $\\var{s} \\ \\mathrm{m}$.
\nc) To find the time taken to reach $\\var{s}m$ we can use $v = u + at$, where $v=0, u= \\var{u}$ and $a = \\var{a}$.
\n\\begin{align} v & = u + at \\\\
0 & = \\var{u} - \\left(\\var{-a} \\times t \\right) \\\\
t & = \\frac{\\var{u}}{\\var{-a}} \\\\
& = \\var{precround(u/-a,3)} \\end{align}
The time taken for the ball to reach $\\var{s} \\ \\mathrm{m}$ from the bottom of the pool is $\\var{precround(u/-a,3)}$ seconds.
d) On the ball's descent back down to the bottom of the pool we can take vertically downwards as the positive direction, and resolve $F=ma$. The resistance of the water still acts against the movement of the ball, but this time the force of gravity is acting in the positive direction.
\n\\begin{align} F & = ma \\\\
\\left( \\var{mass} \\times 9.8 \\right) - \\var{resistance} & = \\var{mass} \\times a \\\\
a & = \\frac{ \\left( \\var{mass} \\times 9.8 \\right) } {\\var{mass}} \\\\
& = \\var{a2} \\end{align}
The acceleration of the ball on its descent back down to its initial position is $\\var{a2} \\ \\mathrm{ms^{-2}}$.
\ne) We have that $a = \\var{a2}$ as the ball heads back down to the bottom of the pool. We can model this journey back down as a particle starting with initial velocity $u=0$ (the ball 'starts' its journey downwards when it reaches its greatest height - where it's speed is zero). We also have the distance it will travel downwards as being $s=\\var{s}$. We want the final velocity $v$ when the ball hits the floor, so we can use $v^2 = u^2 + 2as$.
\n\\begin{align} v^2 & = u^2 + 2as \\\\
v^2 & = 0 + \\left( 2 \\times \\var{a2} \\times \\var{s}\\right) \\\\
& = \\var{2*a2*s}\\\\
v & = \\var{precround((2*a2*s)^(1/2),3)} \\end{align}
The speed of the ball when it hits the bottom of the pool on it's return is $\\var{precround((2*a2*s)^(1/2),3)} \\ \\mathrm{ms^{-1}}$.
\nf) To find the time taken for it to return to the ground from the greatest height of $\\var{s}m$ we can use $s = ut + \\frac{1}{2}at^2$ with $s = \\var{s}, u = 0$ and $a = \\var{a}$.
\n\\begin{align} s & = ut + \\frac{1}{2}at^2 \\\\
\\var{s} &= 0 + \\left( \\frac{1}{2} \\times \\var{a2} \\times t^2 \\right) \\\\
t^2 & = \\frac{2 \\times \\var{s}}{\\var{a2}} \\\\
& = \\var{2*s/a2} \\\\
t & = \\var{precround((2*s/a2)^(1/2),3)} \\end{align}
To find the time taken for the ball to return to the ground from the greatest height is $\\var{precround((2*s/a2)^(1/2),3)}$ seconds.
g) The total time of the ball's journey (to reach the greatest height and fall back down) is found by adding the solutions to part c) and f).
\n\\begin{align} t & = \\var{precround(u/-a,3)} + \\var{precround((2*s/a2)^(1/2),3)} \\\\
& = \\var{precround(u/-a,3) + precround((2*s/a2)^(1/2),3)} \\end{align}
Total time is $\\var{precround(u/-a,3) + precround((2*s/a2)^(1/2),3)}$ seconds.
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