// Numbas version: finer_feedback_settings {"name": "Particle acted on by gravity and a constant force against its motion", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["mass", "acceleration", "acceleration2", "a2", "v_bottom", "max_height", "time_up", "time_down"], "name": "Particle acted on by gravity and a constant force against its motion", "tags": [], "preamble": {"css": "", "js": ""}, "advice": "

a)

We resolve $F=ma$ with the positive direction heading upwards, as the ball is moving upwards. Both the resistance and the force due to gravity are acting in the downwards direction, against the path of the ball.

\\begin{align}
F & = ma \\\\
\\simplify[]{-{resistance} -({mass}*9.8)} &= \\var{mass} \\times a \\\\[0.5em]
a &= \\simplify[]{ (-{resistance}-({mass}*9.8))/{mass} } \\\\[0.5em]
&= \\var{precround(-acceleration,3)}
\\end{align}

We are asked to give the ball's deceleration, which is the opposite of its acceleration. The deceleration of the ball is $\\var{precround(acceleration,3)} \\ \\mathrm{ms^{-2}}$.

b)

When the ball reaches its greatest height it stops moving for an instant before it changes direction. At this instant, $v=0$. Now we have $u = \\var{u}, a = \\var{-precround(acceleration,3)}$ and $v=0$ and we want to find the distance travelled, $s$. We can use the formula $v^2 = u^2 + 2as$, rearranged for $s$.

\\begin{align}
v^2 & = u^2 + 2as \\\\
2as & = v^2 - u^2 \\\\
s & = \\frac{v^2 - u^2}{2a} \\\\[0.5em]
& = \\frac{ 0 - \\var{u}^2}{2 \\times (\\var{-precround(acceleration,3)})} \\\\[0.5em]
& = \\var{precround(max_height,3)}
\\end{align}

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

c)

To find the time taken to reach $\\var{s}\\, \\mathrm{m}$ we can use the equation $v = u + at$, where $v=0, u= \\var{u}$ and $a = \\var{-precround(acceleration,3)}$.

\\begin{align}
v & = u + at \\\\
0 & = \\var{u} - \\left(\\var{a} \\times t \\right) \\\\[0.5em]
t & = \\frac{\\var{u}}{\\var{a}} \\\\[0.5em]
& = \\var{precround(time_up,3)}
\\end{align}

The time taken for the ball to reach $\\var{s} \\, \\mathrm{m}$ from the bottom of the pool is $\\var{precround(time_up,3)}$ 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 ball's movement, but this time the force of gravity acts in the positive direction.

\\begin{align}
F & = ma \\\\
\\simplify[]{{mass}*9.8 - {resistance}} &= \\var{mass} \\times a \\\\[0.5em]
a &= \\simplify[]{({mass}*9.8-{resistance})/{mass}} \\\\[0.5em]
&= \\var{precround(acceleration2,3)}
\\end{align}

The acceleration of the ball on its descent back down to its initial position is $\\var{precround(acceleration2,3)} \\ \\mathrm{ms^{-2}}$.

e)

We have that $a = \\var{precround(acceleration2,3)}$ 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 its speed is zero. We also have the distance it will travel downwards as $s=\\var{precround(max_height,3)}$. We want to know the final velocity $v$ when the ball hits the floor, so we can use the equation $v^2 = u^2 + 2as$.

\\begin{align}
v^2 & = u^2 + 2as \\\\
v^2 & = \\simplify[]{0^2+(2*{precround(acceleration2,3)}*{precround(max_height,3)})} \\\\
& = \\var{2*precround(acceleration2,3)*precround(max_height,3)}\\\\
v & = \\var{precround(v_bottom,3)}
\\end{align}

The speed of the ball when it hits the bottom of the pool on its return is $\\var{precround(v_bottom,3)} \\, \\mathrm{ms^{-1}}$.

f)

To find the time taken for the ball to return to the ground from the greatest height of $\\var{s} \\, \\mathrm{m}$ we can use the equation $s = ut + \\frac{1}{2}at^2$ with $s = \\var{precround(max_height,3)}, u = 0$ and $a = \\var{precround(acceleration2,3)}$.

\\begin{align}
s & = ut + \\frac{1}{2}at^2 \\\\
\\var{precround(max_height,3)} &= \\simplify[]{0 + (1/2)*({precround(acceleration2,3)}*t^2)}\\\\
t^2 & = \\frac{2 \\times \\var{precround(max_height,3)}}{\\var{precround(acceleration2,3)}} \\\\
& = \\var{2*max_height/acceleration2} \\\\
t & = \\var{precround(time_down,3)}
\\end{align}

The ball takes $\\var{precround(time_down,3)}$ seconds to return to the ground from its greatest height.

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 parts c and f.

\\begin{align} t & = \\var{precround(time_up,3)} + \\var{precround(time_down,3)} \\\\
& = \\var{precround(time_up+time_down,3)}
\\end{align}

The total time is $\\var{precround(time_up+time_down,3)}$ seconds.

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Find the vertical deceleration of the ball, in $\\mathrm{ms^{-2}}$ to 3 decimal places.

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Find the greatest height that the ball reaches above the bottom of the pool, in $\\mathrm{m}$ to 3 decimal places.

", "precisionMessage": "