WJEC Unit 4 2024 June — Question 6 8 marks

Exam BoardWJEC
ModuleUnit 4 (Unit 4)
Year2024
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeVector form projectile motion
DifficultyStandard +0.3 This is a straightforward vector projectile motion problem requiring standard SUVAT equations. Part (a) uses constant horizontal velocity to find w, then vertical motion equation to find height. Part (b) finds time to maximum height using v=u+at. All steps are routine applications of mechanics formulas with no conceptual challenges or novel problem-solving required.
Spec1.10h Vectors in kinematics: uniform acceleration in vector form3.02h Motion under gravity: vector form

6. A ball is projected with velocity \(( 4 w \mathbf { i } + 7 w \mathbf { j } ) \mathrm { ms } ^ { - 1 }\) from the top of a vertical tower. After 5 seconds, the ball hits the ground at a point that is 60 m horizontally from the foot of the tower. The unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal and vertical respectively.
  1. Find the value of \(w\) and hence determine the height of the tower.
  2. Determine the proportion of the 5 seconds for which the ball is on its way down.

Question 6:
AnswerMarks
68
Question 6:
6 | 8
6. A ball is projected with velocity $( 4 w \mathbf { i } + 7 w \mathbf { j } ) \mathrm { ms } ^ { - 1 }$ from the top of a vertical tower. After 5 seconds, the ball hits the ground at a point that is 60 m horizontally from the foot of the tower.

The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are horizontal and vertical respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $w$ and hence determine the height of the tower.
\item Determine the proportion of the 5 seconds for which the ball is on its way down.
\end{enumerate}

\hfill \mbox{\textit{WJEC Unit 4 2024 Q6 [8]}}