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This section draws on the skills learnt the previous parts of the 'Differentiation' series of questions, and some logical thinking about the physics of the problem.
\nThe steps within the part walk through the process.
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\nFirstly, differentiate.
\n$\\frac{dy}{dx}=$ [[0]]
\nNow use this result and your knowledge of differentiation to find the maximum height of the missile, rounding to the nearest whole number.
\n$y=$ [[1]]
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\nIt takes the same amount of time to reach its maximum as it does to fall back down.
", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "information"}, {"vsetrangepoints": 5, "prompt": "The maximum height will be at the stationary point.
\nThe stationary point can be found by equating $\\frac{dy}{dx}$ to $0$.
\n$\\frac{dy}{dx}=0=$
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", "allowFractions": true, "variableReplacements": [], "maxValue": "{ct}/(2{cts})", "minValue": "{ct}/(2{cts})", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": true, "showCorrectAnswer": true, "scripts": {}, "marks": "0", "type": "numberentry", "showPrecisionHint": false}], "scripts": {}, "marks": 0, "showCorrectAnswer": true, "type": "gapfill"}], "statement": "An unpowered missile is launched vertically from the ground.
\nAt a time $t$ seconds after the instant of projection, its height, $y$ metres, above the ground is given by the formula
\n\\[ y=\\var{ct}t-\\var{cts}t^2. \\]
\nCalculate the maximum height reached by the missile.
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