Moderate -0.8 This is a straightforward SUVAT application with vertical motion under gravity. Students need to use symmetry (time up = time down = 3s) and apply v = u + at to find initial speed, then use v² = u² + 2as for maximum height. Standard textbook exercise requiring only direct formula application with no problem-solving insight.
3 A particle is thrown vertically upwards and returns to its point of projection after 6 seconds. Air resistance is negligible.
Calculate the speed of projection of the particle and also the maximum height it reaches.
\(0 = u - 9.8 \times 3\), \(u = 29.4\) so \(29.4\) m s\(^{-1}\)
M1, A1
*uvast* leading to \(u\) with \(t=3\) or \(t=6\); gns consistent
\(s = 0.5 \times 9.8 \times 9 = 44.1\) so \(44.1\) m
M1, F1
*uvast* leading to \(s\) with \(t=3\) or \(t=6\) or their \(u\). FT their \(u\) if used with \(t=3\). Signs consistent. Award for 44.1, 132.3 or 176.4 seen. [Award maximum of 3 if one answer wrong]
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = u - 9.8 \times 3$, $u = 29.4$ so $29.4$ m s$^{-1}$ | M1, A1 | *uvast* leading to $u$ with $t=3$ or $t=6$; gns consistent |
| $s = 0.5 \times 9.8 \times 9 = 44.1$ so $44.1$ m | M1, F1 | *uvast* leading to $s$ with $t=3$ or $t=6$ or **their** $u$. FT **their** $u$ if used with $t=3$. Signs consistent. Award for 44.1, 132.3 or 176.4 seen. [Award maximum of 3 if one answer wrong] |
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3 A particle is thrown vertically upwards and returns to its point of projection after 6 seconds. Air resistance is negligible.
Calculate the speed of projection of the particle and also the maximum height it reaches.
\hfill \mbox{\textit{OCR MEI M1 Q3 [4]}}