// Numbas version: exam_results_page_options {"name": "Alan's copy of Numbas demo: Motion under gravity", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Alan's copy of Numbas demo: Motion under gravity", "variables": {"t1": {"definition": "precround(t,2)", "description": "", "name": "t1", "group": "Ungrouped variables", "templateType": "anything"}, "g": {"definition": "9.81", "description": "", "name": "g", "group": "Ungrouped variables", "templateType": "anything"}, "v": {"definition": "random(50..100)", "description": "", "name": "v", "group": "Ungrouped variables", "templateType": "anything"}, "t": {"definition": "V/g", "description": "", "name": "t", "group": "Ungrouped variables", "templateType": "anything"}, "mh": {"definition": "precround(V^2/(2*g),2)", "description": "", "name": "mh", "group": "Ungrouped variables", "templateType": "anything"}}, "variable_groups": [], "advice": "

a)

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Integrating $\\displaystyle{\\frac{\\mathrm{d}^2z}{\\mathrm{d}t^2}=-g}$ once gives the velocity $\\displaystyle{\\frac{\\mathrm{d}z}{\\mathrm{d}t}=-gt+A}$.

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But $A=\\var{v}$ as the velocity is $\\var{V}\\;m/s$ at $t=0$.

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So the velocity is

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\\begin{align} \\frac{\\mathrm{d}z}{\\mathrm{d}t} &= \\var{v}-gt & (1) \\end{align}

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Integrating again gives

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\\[ z = \\var{v}t-\\frac{g}{2}t^2+B \\]

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and $B=0$ as $z=0$ at $t=0$.

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Hence the distance travelled upwards is given by

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\\begin{align} z &= \\var{v}t-\\frac{g}{2}t^2 & (2) \\end{align}

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{advicegraph()}

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b)

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The time $t_{\\text{max}}$ taken to reach maximum height is the time satisfying $\\displaystyle{\\frac{dz}{dt}=0}$

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$t_{\\text{max}}$ is given from equation $(1)$ by

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\\begin{align}
\\var{v} - gt_{\\text{max}} &= 0 \\\\
gt_{\\text{max}} &= \\var{v} \\\\
t_{\\text{max}} &= \\frac{\\var{v}}{g} \\\\[0.5em]
&= \\frac{\\var{v}}{9.81} \\\\[0.5em]
&= \\var{t1}
\\end{align}

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(to $2$ decimal places)

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This is at the point $A$ in the graph above.

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c)

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The maximum height $z_{\\text{max}}$ is given from equation $(2)$ by substituting in the value $t_{\\text{max}}= \\var{v}/g$, giving

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\\begin{align}
z_{\\text{max}} &= \\var{v} \\times \\frac{\\var{v}}{g} - \\frac{g}{2}\\left(\\frac{\\var{v}}{g}\\right)^2 \\\\
&= \\frac{\\var{v}^2}{g}-\\frac{g\\var{v}^2}{2g^2} \\\\
&= \\frac{\\var{v}^2}{2g} \\\\
&= \\var{mh}\\;m
\\end{align}

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(to $2$ decimal places)

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This is at the point $B$ in the graph above.

", "ungrouped_variables": ["t", "v", "mh", "g", "t1"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "statement": "

A Numbas question can include interactive graphics, such as this plot of the trajectory given by the student's answer.

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See this question in the public editor

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A ball is thrown upwards, and moves according to the equation $\\displaystyle{\\frac{d^2z}{dt^2}=-g}$
(where $z(t)$ is distance in metres measured upwards from the ground and the constant acceleration of gravity, $g$ , is given as $9.81\\;m/s^2$).

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The ball is projected upwards with a speed $\\var{v}\\;m/s$.

", "metadata": {"description": "

Customised for the Numbas demo exam

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Motion under gravity. Object is projected vertically with initial velocity $V\\;m/s$. Find time to maximum height and the maximum height. Now includes an interactive plot.

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{graphsolution()}

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Input the vertical distance $z$ as a  function of $t$.

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Note that at $t=0$ we have $z=0$ and that $\\displaystyle \\frac{dz}{dt}=\\var{v}m/s$.

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Input gravitational acceleration as $g$.

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$z=$ [[0]]

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Your formula is plotted in the graph above. The vertical axis represents $z$ and the horizontal axis represents $t$.

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Note that the blue line indicates that:

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    \n
  1. Your solution should go through $(0,0)$;
  2. \n
  3. Your solution should have this line as the tangent to the curve at $(0,0)$, because $\\displaystyle \\frac{\\mathrm{d}z}{\\mathrm{d}t}=\\var{v}\\; m/s$.
  4. \n
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Time taken to reach maximum height = [[0]] $s$ (accurate to $2$ decimal places)

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Maximum height = [[1]] $m$ (accurate to $2$ decimal places)

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