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From the steps provided, we found that at the stationary point of the projectile parabola:

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$\\frac{dy}{dx}=0$    and    $t=\\simplify{{ct}/{2*{cts}}}$

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To find the maximum height, the value for $t$ at the stationary point is substituted back into the original equation:

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$y=\\var{ct}t-\\var{cts}t^2=\\var{ct}\\left(\\simplify{{ct}/{2*{cts}}}\\right)-\\var{cts}\\left(\\simplify{{ct}/{2*{cts}}}\\right)^2$

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$y=\\var{ct}t-\\var{cts}t^2$

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Firstly, differentiate.

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$\\frac{dy}{dx}=$ [[0]]

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Now use this result and your knowledge of differentiation to find the maximum height of the missile, rounding to the nearest whole number.

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$y=$ [[1]]

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The missile will follow a parabolic curve when height is plotted against time.

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It takes the same amount of time to reach its maximum as it does to fall back down.

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The maximum height will be at the stationary point.

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The stationary point can be found by equating $\\frac{dy}{dx}$ to $0$.

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$\\frac{dy}{dx}=0=$

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Therefore, $t=$

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An unpowered missile is launched vertically from the ground.

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At a time $t$ seconds after the instant of projection, its height, $y$ metres, above the ground is given by the formula

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\\[ y=\\var{ct}t-\\var{cts}t^2. \\]

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Calculate the maximum height reached by the missile.

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coeff of t^2

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coeff of t

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Real life problems with differentiation

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