Mathematics 9709/43 — October/November 2023

Walkthrough

The particle is projected upwards, so we choose upwards as the positive direction. Gravity acts downwards, so the acceleration is $a = -10\text{ m s}^{-2}$ (using $g = 10$). The standard constant-acceleration formula is $s = ut + \frac{1}{2}at^2$. We apply it at $t = 3$ and at $t = 4$. At both times the height is the same value $s$, so we can set the two expressions equal. This gives a linear equation in $u$ only. Solving it gives $u = 35$. Substituting this value back into either expression gives $s = 60$.

An alternative is to notice that the times 3 and 4 are symmetric about the time of maximum height, 3.5 seconds. At maximum height the velocity is zero, so using $v = u + at$ gives $0 = u - 10(3.5)$, hence $u = 35$. Then substitute into $s = ut + \frac{1}{2}at^2$ at $t = 3$ to get $s = 60$.

Key Takeaways

This question tests the use of the suvat equations for motion with constant acceleration. The key idea is that the same height at two different times can be used to form two equations and solve for the initial speed. It also reinforces the importance of a consistent sign convention: upwards positive means gravity is negative.

Common Mistakes

• Forgetting to include the negative sign for gravity. If you write $s = ut + 5t^2$, you get the wrong equations.
• Not equating the two expressions for $s$, so you never eliminate $s$.
• Assuming the particle is at its maximum height at $t = 3$ or $t = 4$. It is not; the maximum occurs halfway between, at $t = 3.5$.
• Giving an unsupported answer. The mark scheme requires the suvat equations to be shown (M1/A1), so a numerical answer with no working may not earn full marks.

Things to Be Careful About

• Use $g = 10\text{ m s}^{-2}$ unless the question states otherwise; this mark scheme uses $g = 10$.
• Keep the sign convention consistent throughout. Taking upwards as positive makes $a = -10$, while $u$ and $s$ are positive.
• Check units: $u$ is in $\text{m s}^{-1}$ and $s$ is in metres.
• When using the alternative maximum-height method, remember that the maximum height occurs at the average of the two times, $t = 3.5$, not at either given time.