A particle $P$ with mass $0.2\,\text{kg}$ is let go from rest at a point $O$ above horizontal ground. At time $t\,\text{s}$ after release, the downward velocity of $P$ is $v\,\text{m s}^{-1}$. A force of magnitude $0.6t\,\text{N}$ acts vertically downwards on $P$. A vertically upwards force of magnitude $ke^{-t}\,\text{N}$, where $k$ is a constant, also acts on $P$.
(i)[2]
Show that $\dfrac{dv}{dt}$ can be written as $10 - 5ke^{-t} + 3t$.
(ii)[3]
Find the largest value of $k$ such that $P$ does not initially move upwards.
(iii)[5]
Given that $k = 1$, and that $P$ hits the ground at $t = 2$, determine the height of $O$ above the ground.
(c(iii))[5]
Given that $k = 1$, and that $P$ strikes the ground at $t = 2$, find how high $O$ is above the ground.
Worked solution & mark scheme
This 15-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Use Newton’s Second Law downwards, giving $0.2 \frac{dv}{dt} = 0.2 g + 0.6 t - k e^{-t}$” …