2015 A-level H2 Mathematics (9740) Paper 2 Question 1 Suggested Solutions

By KS Teng

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
16 -\frac{1}{2} h \ge 0

\Rightarrow h \le 32

\therefore, \mathrm{maximum~h}=32m

(ii)
\frac{dh}{dt} = \frac{1}{10} \frac{\sqrt{32-h}}{\sqrt{2}}

\int \frac{1}{\sqrt{32-h}} ~dh = \int \frac{1}{10 \sqrt{2}}~dt

-2 \sqrt{32-h} + C = \frac{1}{10 \sqrt{2}}t

t = - 20 \sqrt{64-2h} + 10 \sqrt{2}C

When t = 0, h = 0, C = \frac{16}{\sqrt{2}}

\therefore, t = - 20 \sqrt{64-2h} + 160

When h = 16m, t = 46.9 years.

KS Comments:

Straightforward question.

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    2011 A-level H2 Mathematics (9740) Paper 2 Question 9 Suggested SolutionsJC Mathematics
    2015 A-level H2 Mathematics (9740) Paper 2 Question 2 Suggested SolutionsJC Mathematics