Moderate -0.8 This is a straightforward two-stage kinematics problem using standard SUVAT equations. Students apply v² = u² + 2as to find V, then use v = u + at twice (downward motion and rebound). All steps are routine applications of memorized formulas with no problem-solving insight required, making it easier than average but not trivial due to the two-stage calculation.
A small ball is projected vertically downwards with speed \(5\text{ m s}^{-1}\) from a point \(A\) at a height of \(7.2\text{ m}\) above horizontal ground. The ball hits the ground with speed \(V\text{ m s}^{-1}\) and rebounds vertically upwards with speed \(\frac{1}{2}V\text{ m s}^{-1}\). The highest point the ball reaches after rebounding is \(B\). Find \(V\) and hence find the total time taken for the ball to reach the ground from \(A\) and rebound to \(B\). [5]
Question 2:
2 | [V2 =52+2×g×7.2] | M1 | Use of uvast to find V
V =13 | A1
[13 = 5 + gt t = ….. ] 0.8 (s) | M1 | Use of uvast to find time for A to reach ground
[0 = 6.5 – gt t = ….. ] 0.65 (s) | M1 | Use of uvast to find time from ground to B
Total time is 1.45 (s) | A1
Total: | 5
Question | Answer | Marks | Guidance
A small ball is projected vertically downwards with speed $5\text{ m s}^{-1}$ from a point $A$ at a height of $7.2\text{ m}$ above horizontal ground. The ball hits the ground with speed $V\text{ m s}^{-1}$ and rebounds vertically upwards with speed $\frac{1}{2}V\text{ m s}^{-1}$. The highest point the ball reaches after rebounding is $B$. Find $V$ and hence find the total time taken for the ball to reach the ground from $A$ and rebound to $B$. [5]
\hfill \mbox{\textit{CAIE M1 2018 Q2 [5]}}