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From the steps provided, we found that at the stationary point of the projectile parabola:
\n$\\frac{dy}{dx}=0$ and $t=\\simplify{{ct}/{2*{cts}}}$
\nTo find the maximum height, the value for $t$ at the stationary point is substituted back into the original equation:
\n$y=\\var{ct}t-\\var{cts}t^2=\\var{ct}\\left(\\simplify{{ct}/{2*{cts}}}\\right)-\\var{cts}\\left(\\simplify{{ct}/{2*{cts}}}\\right)^2$
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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]]
\n", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "steps": [{"prompt": "The missile will follow a parabolic curve when height is plotted against time.
\nIt takes the same amount of time to reach its maximum as it does to fall back down.
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\nThe stationary point can be found by equating $\\frac{dy}{dx}$ to $0$.
\n$\\frac{dy}{dx}=0=$
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\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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