Pre-U Pre-U 9794/3 2014 June — Question 7 5 marks

Exam BoardPre-U
ModulePre-U 9794/3 (Pre-U Mathematics Paper 3)
Year2014
SessionJune
Marks5
TopicSUVAT in 2D & Gravity
TypeVertical projection: max height
DifficultyEasy -1.2 This is a straightforward kinematics problem using constant acceleration equations (SUVAT). Students need to apply v² = u² + 2as for maximum height and s = ut + ½at² for time of flight—both standard textbook exercises requiring only direct substitution with no problem-solving insight or multi-step reasoning beyond basic algebraic manipulation.
Spec3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form
A stone is projected vertically upwards from ground level at a speed of \(30\,\mathrm{m}\,\mathrm{s}^{-1}\). It is assumed that there is no wind or air resistance. Find the maximum height it reaches and the total time it takes from its projection to its return to ground level. [5]
AnswerMarks Guidance
At max height
\(0 = 30^2 - 2 \times 10 \times h\)M1 A1 Use of an appropriate 'suvat' equation. Correct equation.
\(\therefore h = 45\) mA1 Correct outcome. Allow \(g = 9.8\), giving \(h = 45.918\)
On return to ground level
\(-30 = 30 - 10 \times t\)M1 Correct use of a second appropriate 'suvat' equation. Allow any valid method, e.g. (time to max ht) × 2.
\(\therefore t = 6\) secA1 Correct outcome. Allow \(g = 9.8\), giving \(t = 6.122\) [5] 
At max height | 

$0 = 30^2 - 2 \times 10 \times h$ | M1 A1 | Use of an appropriate 'suvat' equation. Correct equation.

$\therefore h = 45$ m | A1 | Correct outcome. Allow $g = 9.8$, giving $h = 45.918$

On return to ground level | 

$-30 = 30 - 10 \times t$ | M1 | Correct use of a second appropriate 'suvat' equation. Allow any valid method, e.g. (time to max ht) × 2.

$\therefore t = 6$ sec | A1 | Correct outcome. Allow $g = 9.8$, giving $t = 6.122$ [5]
A stone is projected vertically upwards from ground level at a speed of $30\,\mathrm{m}\,\mathrm{s}^{-1}$. It is assumed that there is no wind or air resistance. Find the maximum height it reaches and the total time it takes from its projection to its return to ground level. [5]

\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2014 Q7 [5]}}
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