A particle is projected vertically upwards with speed $$\(40 \mathrm{~m} \mathrm{~s}^{-1}\)$$ alongside a building of height $$\(h \mathrm{~m}\)$$. One second after the first particle is projected, a second particle is projected vertically upwards from the top of the building with speed $$\(20 \mathrm{~m} \mathrm{~s}^{-1}\)$$. Denoting the time after projection of the first particle by $$\(t \mathrm{~s}\)$$, find the value of $$\(t\)$$ for which the two particles are at the same height above the ground. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_w20_qp_42 Year：2020 Question No：5(b)
Answer:
Height of first particle above ground $=40 t-1 / 2 \times 10 t^{2}$
Height of second particle above top of building
$
=20(t-1)-1 / 2 \times 10 \times(t-1)^{2}
$
$
60+20(t-1)-1 / 2 \times 10 \times(t-1)^{2}=40 t-1 / 2 \times 10 t^{2}
$$
t=3.5 \mathrm{~seconds}
$
Alternative method for question 5(b)
$
\begin{array}{l}
h_{1}=40 \times 1-5 \times 1^{2}[=35] \text { and } v_{1}=40-10 \times 1[=30] \\
H_{1}=30 T-5 \times T^{2}, H_{2}=20 T-5 \times T^{2}
\end{array}
$$
30 T-5 \times T^{2}=20 T-5 \times T^{2}+(60-35)
$$
T=2.5 \text { and so time to meet }=2.5+1=3.5 \text { seconds }
$
Height of second particle above top of building
$
=20(t-1)-1 / 2 \times 10 \times(t-1)^{2}
$
$
60+20(t-1)-1 / 2 \times 10 \times(t-1)^{2}=40 t-1 / 2 \times 10 t^{2}
$$
t=3.5 \mathrm{~seconds}
$
Alternative method for question 5(b)
$
\begin{array}{l}
h_{1}=40 \times 1-5 \times 1^{2}[=35] \text { and } v_{1}=40-10 \times 1[=30] \\
H_{1}=30 T-5 \times T^{2}, H_{2}=20 T-5 \times T^{2}
\end{array}
$$
30 T-5 \times T^{2}=20 T-5 \times T^{2}+(60-35)
$$
T=2.5 \text { and so time to meet }=2.5+1=3.5 \text { seconds }
$
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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