Edexcel M1 2017 October — Question 5 11 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2017
SessionOctober
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeVertical motion under gravity
DifficultyModerate -0.8 This is a straightforward three-part SUVAT question requiring standard application of kinematic equations with constant acceleration. Part (a) uses v²=u²+2as to find maximum height, part (b) requires solving a quadratic from s=ut+½at², and part (c) again uses v²=u²+2as. All parts are routine textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the multi-step nature and need for careful sign conventions.
Spec3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form

A small ball is projected vertically upwards from a point \(O\) with speed 14.7 m s\(^{-1}\). The point \(O\) is 2.5 m above the ground. The motion of the ball is modelled as that of a particle moving freely under gravity. Find
  1. the maximum height above the ground reached by the ball, [4]
  2. the time taken for the ball to first reach a height of 1 m above the ground, [4]
  3. the speed of the ball at the instant before it strikes the ground for the first time. [3]

A small ball is projected vertically upwards from a point $O$ with speed 14.7 m s$^{-1}$. The point $O$ is 2.5 m above the ground. The motion of the ball is modelled as that of a particle moving freely under gravity.

Find

\begin{enumerate}[label=(\alph*)]
\item the maximum height above the ground reached by the ball,
[4]
\item the time taken for the ball to first reach a height of 1 m above the ground,
[4]
\item the speed of the ball at the instant before it strikes the ground for the first time.
[3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2017 Q5 [11]}}