A particle is projected vertically upwards with speed $$\(u \mathrm{~m} \mathrm{~s}^{-1}\)$$ from a point on horizontal ground. After 2 seconds, the height of the particle above the ground is $$\(24 \mathrm{~m}\)$$. The height of the particle above the ground is more than $$\(h \mathrm{~m}\)$$ for a period of $$\(3.6 \mathrm{~s}\)$$. Find $$\(h\)$$ .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_s21_qp_43 Year：2021 Question No：4(b)
Answer:
At maximum height \(0=22^{2}-2 g s\)
Maximum height \(s=24.2 \mathrm{~m}\)
Height down \(=0.5 g \times 1.8^{2}(=16.2)\)
\(h=8\)
Alternative method for Question 4 (b)
\( 0=22-10 t
\)
\(t=2.2\)
\( h=22 \times(2.2-1.8)-\frac{1}{2} g \times(2.2-1.8)^{2}
\)
\( h=8
\)
Alternative method for Question 4(b)
\( 22 \mathrm{t}-\frac{1}{2} g t^{2}=22 \times(t+3.6)-\frac{1}{2} g \times(t+3.6)^{2}
\)
\(t=0.4(\) or \(t+3.6=4)\)
\(h=22 \times 0.4-\frac{1}{2} g \times 0.4^{2}\)
\( h=8
\)
Maximum height \(s=24.2 \mathrm{~m}\)
Height down \(=0.5 g \times 1.8^{2}(=16.2)\)
\(h=8\)
Alternative method for Question 4 (b)
\( 0=22-10 t
\)
\(t=2.2\)
\( h=22 \times(2.2-1.8)-\frac{1}{2} g \times(2.2-1.8)^{2}
\)
\( h=8
\)
Alternative method for Question 4(b)
\( 22 \mathrm{t}-\frac{1}{2} g t^{2}=22 \times(t+3.6)-\frac{1}{2} g \times(t+3.6)^{2}
\)
\(t=0.4(\) or \(t+3.6=4)\)
\(h=22 \times 0.4-\frac{1}{2} g \times 0.4^{2}\)
\( h=8
\)
Knowledge points:
4.2.4 use appropriate formulae for motion with constant acceleration in a straight line. （Questions may involve setting up more than one equation, using information about the motion of different particles.）
4.4.3 solve simple problems which may be modelled as the motion of a particle moving vertically or on an inclined plane with constant acceleration Including, for example, motion of a particle on a rough plane where the acceleration while moving up the plane is different from the acceleration while moving down the plane.
Solution:
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