// Numbas version: exam_results_page_options {"name": "Ethan'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": [{"metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Customised for the Numbas demo exam

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.

"}, "statement": "

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

See this question in the public editor

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 $9.81\\;m/s^2$).

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

", "ungrouped_variables": ["t", "v", "mh", "g", "t1"], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"mh": {"group": "Ungrouped variables", "templateType": "anything", "name": "mh", "description": "", "definition": "precround(V^2/(2*g),2)"}, "v": {"group": "Ungrouped variables", "templateType": "anything", "name": "v", "description": "", "definition": "random(50..100)"}, "g": {"group": "Ungrouped variables", "templateType": "anything", "name": "g", "description": "", "definition": "9.81"}, "t": {"group": "Ungrouped variables", "templateType": "anything", "name": "t", "description": "", "definition": "V/g"}, "t1": {"group": "Ungrouped variables", "templateType": "anything", "name": "t1", "description": "", "definition": "precround(t,2)"}}, "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "variable_groups": [], "parts": [{"type": "gapfill", "prompt": "

{graphsolution()}

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

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

Input gravitational acceleration as $g$.

$z=$ [[0]]

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

Note that the blue line indicates that:

Your solution should go through $(0,0)$;
Your solution should have this line as the tangent to the curve at $(0,0)$, because $\\displaystyle \\frac{\\mathrm{d}z}{\\mathrm{d}t}=\\var{v}\\; m/s$.

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Time taken to reach maximum height = [[0]] $s$ (accurate to $2$ decimal places)

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

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a)

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

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

So the velocity is

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

Integrating again gives

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

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

Hence the distance travelled upwards is given by

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

{advicegraph()}

b)

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

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

\\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}

(to $2$ decimal places)

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

c)

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

\\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}

(to $2$ decimal places)

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

", "name": "Ethan's copy of Numbas demo: Motion under gravity", "extensions": ["jsxgraph"], "type": "question", "contributors": [{"name": "Ethan Smith", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/739/"}]}]}], "contributors": [{"name": "Ethan Smith", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/739/"}]}