(i)[2]
Show that, as $P$ travels upwards, $v\,\frac{dv}{dx} = -10 - 0.04v^2$.
(ii)[3]
Find the greatest height reached by $P$ above the surface.
(iii)[4]
Find the speed of $P$ just before it strikes the surface on the way down.
Mathematics 9709 · AS & A Level · Probability
Probability — practice question
A particle $P$ with mass $0.5\,\text{kg}$ is launched vertically upwards from a point on a horizontal surface. A resisting force of magnitude $0.02v^2\,\text{N}$ acts on $P$, where $v\,\text{m s}^{-1}$ is the upward speed of $P$ when it is at a height of $x\,\text{m}$ above the surface. Its initial speed is $8\,\text{m s}^{-1}$.