Moderate -0.8 This is a straightforward SUVAT application with standard values. Students need to identify the correct kinematic equation (s = ut + ½at²), substitute given values (s = -2, u = 3.5, a = -9.8), and solve the resulting quadratic. It's more routine than average A-level questions since it requires only one equation and basic algebraic manipulation, though the quadratic solving adds minor complexity.
10 A small ball \(B\) is projected vertically upwards from a point 2 m above horizontal ground. \(B\) is projected with initial speed \(3.5 \mathrm {~ms} ^ { - 1 }\), and takes \(t\) seconds to reach the ground.
Find the value of \(t\).
Correct equation – need not be simplified; A1 all equations correct – need not be simplified
\(t=1.09\)
A1
BC (positive root only); \(t=1.0890679\ldots\)
## Question 10:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s=ut+\frac{1}{2}at^2$ with $s=\pm2$, $u=\pm3.5$, $a=\pm g$ and $t$ is the total time for the motion | M1 | Possible to consider eg motion up and motion down; M1 for **complete** method to find relevant times |
| $-2=3.5t+\frac{1}{2}(-9.8)t^2 \Rightarrow 4.9t^2-3.5t-2=0$ | A1 | Correct equation – need not be simplified; A1 all equations correct – need not be simplified |
| $t=1.09$ | A1 | **BC** (positive root only); $t=1.0890679\ldots$ |
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10 A small ball $B$ is projected vertically upwards from a point 2 m above horizontal ground. $B$ is projected with initial speed $3.5 \mathrm {~ms} ^ { - 1 }$, and takes $t$ seconds to reach the ground.
Find the value of $t$.
\hfill \mbox{\textit{OCR PURE Q10 [3]}}