Easy -1.2 This is a straightforward SUVAT application requiring students to set up s = ut + ½at² with known values (u=8, s=3, a=-9.8) and solve the resulting quadratic equation. It's a standard textbook exercise with clear setup and routine algebraic manipulation, making it easier than average for A-level.
1 A ball is thrown vertically upwards with a speed of \(8 \mathrm {~ms} ^ { - 1 }\).
Find the times at which the ball is 3 m above the point of projection.
Using *suvat* with upwards positive; \(u=8, a=-9.8, s=3\)
M1
Using *suvat* equation(s) leading to a value for \(t\). Allow sign errors
\(s = ut + \frac{1}{2}at^2\)
\(3 = 8t - 4.9t^2\)
Solve \(4.9t^2 - 8t + 3 = 0\)
\(t = 0.584, 1.05\) s
A1
Both values needed. ISW if an inequality is given as the final answer. Exact roots are \(\frac{40 \pm \sqrt{130}}{49}\)
Total: [2]
## Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using *suvat* with upwards positive; $u=8, a=-9.8, s=3$ | M1 | Using *suvat* equation(s) leading to a value for $t$. Allow sign errors |
| $s = ut + \frac{1}{2}at^2$ | | |
| $3 = 8t - 4.9t^2$ | | |
| Solve $4.9t^2 - 8t + 3 = 0$ | | |
| $t = 0.584, 1.05$ s | A1 | Both values needed. ISW if an inequality is given as the final answer. Exact roots are $\frac{40 \pm \sqrt{130}}{49}$ |
**Total: [2]**
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1 A ball is thrown vertically upwards with a speed of $8 \mathrm {~ms} ^ { - 1 }$.\\
Find the times at which the ball is 3 m above the point of projection.
\hfill \mbox{\textit{OCR MEI Paper 1 2023 Q1 [2]}}