Find the initial speed in $\\mathrm{ms^{-1}}$ (to 3 decimal places) of the ball when it is thrown.
"}, {"precisionType": "dp", "mustBeReduced": false, "strictPrecision": false, "variableReplacementStrategy": "originalfirst", "type": "numberentry", "customMarkingAlgorithm": "", "scripts": {}, "mustBeReducedPC": 0, "precision": "3", "showFeedbackIcon": true, "minValue": "t1-t2-0.02", "showPrecisionHint": false, "variableReplacements": [], "precisionPartialCredit": 0, "showCorrectAnswer": true, "unitTests": [], "maxValue": "t1-t2+0.02", "allowFractions": false, "marks": 1, "correctAnswerStyle": "plain", "correctAnswerFraction": false, "precisionMessage": "You have not given your answer to the correct precision.", "extendBaseMarkingAlgorithm": true, "notationStyles": ["plain", "en", "si-en"], "prompt": "Find the total time in $\\mathrm{s}$ for which the ball is $\\var{s-(15+s2)} \\mathrm{m}$ or more above $X$.
\nGive your answer to 3 decimal places.
"}], "preamble": {"css": "", "js": ""}, "variables": {"u": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "definition": "precround((2*9.8*s)^(1/2),3)", "name": "u"}, "t1": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "definition": "precround((u+(u^2-19.6*distance)^(1/2))/9.8,4)", "name": "t1"}, "s2": {"group": "Ungrouped variables", "description": "", "templateType": "randrange", "definition": "random(0..10#1)", "name": "s2"}, "s": {"group": "Ungrouped variables", "description": "greatest height reached by the ball
", "templateType": "randrange", "definition": "random(40..80#1)", "name": "s"}, "distance": {"group": "Ungrouped variables", "description": "total time for which the ball is distance or more above X
", "templateType": "anything", "definition": "s-(15+s2)", "name": "distance"}, "a": {"group": "Ungrouped variables", "description": "", "templateType": "number", "definition": "-9.8", "name": "a"}, "t2": {"group": "Ungrouped variables", "description": "", "templateType": "anything", "definition": "precround((u-(u^2-19.6*distance)^(1/2))/9.8,4)", "name": "t2"}}, "rulesets": {}, "functions": {}, "ungrouped_variables": ["s", "a", "s2", "distance", "u", "t1", "t2"], "statement": "A ball is thrown vertically upwards from a point $X$. The greatest height reached by the ball is $\\var{s} \\mathrm{m}$ above $X$. The acceleration due to gravity is $g = 9.8 \\mathrm{ms^{-2}}$.
", "variable_groups": [], "tags": [], "variablesTest": {"condition": "distance>0", "maxRuns": 100}, "advice": "We are told that the ball is thrown upwards, so let the upwards direction be positive and gravity will act in the opposite direction, thus $a=-9.8$. At the ball's greatest height, $s=\\var{s}$, the velocity will be $v=0$ for an instant as the ball changes direction and falls downwards. We are trying to find the initial velocity of the ball, $u$. We have everything except $t$ so we can use the formula $v^2 = u^2 + 2as$, rearranged for $u$.
\n\\begin{align} v^2 & = u^2 + 2as, \\\\
0 & = u^2 + \\left(2 \\times \\var{a} \\times \\var{s}\\right), \\\\
0 & = u^2 - \\var{2*-a*s}.
\\end{align}
Therefore
\n\\begin{align} u^2 &= \\var{2*-a*s}, \\\\
u & = \\var{precround ((2*-a*s)^(1/2),3)} \\mathrm{ms^{-1}}.
\\end{align}
The initial velocity of the ball is $ \\var{precround ((2*-a*s)^(1/2),3)} \\mathrm{ms^{-1}}.$
\nWe have now that $s=\\var{distance}$, $u = \\var{u}$ and $a=\\var{a}$. To find the time the ball is above $\\var{distance} \\mathrm{m}$ we need to find the time the ball first passes over $\\var{distance} \\mathrm{m}$ and the time it then falls back down past $\\var{distance} \\mathrm{m}$, and find the difference between those two times. To find these times we can use the formula $s=ut+\\frac{1}{2}at^2$, rearranged for $t$.
\nWe have
\n\\begin{align} s & = ut + \\frac{1}{2}at^2, \\\\
\\var{distance} & = \\var{u}t - \\left( \\frac{1}{2} \\times \\var{-a} \\times t^2 \\right), \\\\
4.9t^2 - \\var{u}t + \\var{distance} & = 0.
\\end{align}
Therefore our first $t$ value, when the ball first passes $\\var{distance} \\mathrm{m}$, is
\n\\begin{align} t & = \\frac{\\var{u} + \\sqrt{\\var{u}^2 - \\left(4 \\times 4.9 \\times \\var{distance} \\right)} }{2 \\times 4.9}, \\\\
& = \\var{t1} \\mathrm{s},
\\end{align}
and our second $t$ value, when the ball second passes $\\var{distance} \\mathrm{m}$, is
\n\\begin{align} t & = \\frac{\\var{u} - \\sqrt{\\var{u}^2 - \\left(4 \\times 4.9 \\times \\var{distance} \\right)} }{2 \\times 4.9}, \\\\
& = \\var{t2} \\mathrm{s}.
\\end{align}
Then the total time for which the ball is $\\var{distance} \\mathrm{m}$ or more above $X$ is $\\var{t1} - \\var{t2} = \\var{precround(t1-t2,3)}$ seconds.
", "extensions": [], "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "name": "Simon's copy of SUVAT equations question 14", "type": "question", "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}]}], "contributors": [{"name": "Amy Chadwick", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/505/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}