Easy -1.2 This is a straightforward one-dimensional kinematics problem requiring only substitution into SUVAT equations. With 3 marks, it involves setting up s = ut + ½at² with known values and solving a quadratic equation—routine mechanics with no conceptual difficulty or problem-solving insight required.
A particle \(P\) is projected vertically upwards with speed \(24 \text{ m s}^{-1}\) from a point \(5 \text{ m}\) above ground level. Find the time from projection until \(P\) reaches the ground. [3]
Question 1:
1 | –5 = 24t – 5t 2 | M1 | 1
Use s = ut + at 2
2
5t 2 – 24t – 5 = 0 | M1 | Solve relevant 3 term quadratic
t = 5 | A1
3
Alternative scheme for Question 1
0 = 24 – 10t → t = 2.4
1 1 | M1 | Attempt to find the time taken to reach the
highest point
0 = 242 + 2 × (–10) × h → h = 28.8
1
And 33.8 = gt 2 → t = 2.6
2 2
2 | M1 | Find total height h reached and attempt to
find time taken from highest point to
ground level
t = t + t = 5
1 2 | A1
Question | Answer | Marks | Guidance
A particle $P$ is projected vertically upwards with speed $24 \text{ m s}^{-1}$ from a point $5 \text{ m}$ above ground level. Find the time from projection until $P$ reaches the ground. [3]
\hfill \mbox{\textit{CAIE M1 2018 Q1 [3]}}