2022TrialShort answer

Q14 (c) (i)

1 mark

Question

A $20\,\mathrm{kg}$ projectile is launched upwards from the ground at $500\,\mathrm{m/s}$ . It experiences a gravitational force of $200\,\mathrm N$ and air resistance, opposite to its motion, of $\frac{v^2}{18}\,\mathrm N$ . Letting the ground be $x=0$ , show that the acceleration is
$\ddot{x}=-10-\frac{v^2}{360}.$

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Solution

Q.14(c)(i)

\downarrow 200\,\mathrm{N}\qquad \downarrow\left(\frac{v^2}{18}\,\mathrm{N}\right)

F=ma=-200-\frac{v^2}{18}

20\ddot{x}=-200-\frac{v^2}{18}

\ddot{x}=a=-10-\frac{v^2}{360}.

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Syllabus alignment

Mechanics

Resisted motionDerive and solve motion models with resistance on horizontal and vertical paths, including terminal velocity.

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