| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: time at height |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT question requiring standard application of kinematic equations with gravity. Part (a) uses symmetry of vertical motion and s=ut+½at², part (b) uses v²=u²+2as at maximum height, and part (c) solves a quadratic equation. All techniques are routine M1 exercises with no problem-solving insight required, making it easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
\begin{enumerate}
\item A stone is projected vertically upwards from a point $A$ with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$. After projection the stone moves freely under gravity until it returns to $A$. The time between the instant that the stone is projected and the instant that it returns to $A$ is $3 \frac { 4 } { 7 }$ seconds.
\end{enumerate}
Modelling the stone as a particle,\\
(a) show that $u = 17 \frac { 1 } { 2 }$,\\
(b) find the greatest height above $A$ reached by the stone,\\
(c) find the length of time for which the stone is at least $6 \frac { 3 } { 5 } \mathrm {~m}$ above $A$.\\
\hfill \mbox{\textit{Edexcel M1 2012 Q5 [11]}}