| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: max height |
| Difficulty | Easy -1.2 This is a straightforward SUVAT application with standard vertical projection under gravity. Part (i) uses v²=u²+2as to find maximum height (where v=0), and part (ii) uses s=ut+½at² to find time when s=-5m. Both are routine textbook exercises requiring only direct formula substitution with no problem-solving insight needed. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
**Question 8(i)**
Attempt to use $v^2 = u^2 + 2as$ with $v = 0$ and $a = -g$ to find height **M1**
Get $12.2\ \text{m}$ **A1** (7.2 gets M1A0)
**Question 8(ii)**
Set up quadratic equation in $t$ $[-5 = 12t - 5t^2]$ **M1** May be done in 2 parts. M1 for one part and M1 for the second and added to first
Attempt to solve their 3 term quadratic **M1**
Get $t = 2.76\ \text{s}$ **A1**8 A small ball is thrown vertically upwards with speed $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, from a point 5 m above the ground. Assuming air resistance is negligible, find\\
(i) the greatest height above the ground reached by the ball,\\
(ii) the time taken for the ball to reach the ground.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2018 Q8 [7]}}