A particle is projected vertically upwards with speed $$\(40 \mathrm{~m} \mathrm{~s}^{-1}\)$$ alongside a building of height $$\(h \mathrm{~m}\)$$. Given that the particle is above the level of the top of the building for $$\(4 \mathrm{~s}\)$$, find $$\(h\)$$. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_w20_qp_42 Year：2020 Question No：5(a)
Answer:
$
40-g t=0 \quad[t=4]
$
Time to top of building \(=4-1 / 2(4)=2\)
$
\begin{array}{l}
h=40 \times 2-1 / 2 \times 10 \times 2^{2}
h=40 \times 6-1 / 2 \times 10 \times 6^{2}
\end{array}
$
$
h=60
$
Alternative method for question \(5(\) a)
$
0=40^{2}+2 \times(-10) \times H
$
$
H=80
$
$
s=1 / 2 \times 10 \times 2^{2}
$
$
s=20 \text { and so } h=80-20=60
$
40-g t=0 \quad[t=4]
$
Time to top of building \(=4-1 / 2(4)=2\)
$
\begin{array}{l}
h=40 \times 2-1 / 2 \times 10 \times 2^{2}
h=40 \times 6-1 / 2 \times 10 \times 6^{2}
\end{array}
$
$
h=60
$
Alternative method for question \(5(\) a)
$
0=40^{2}+2 \times(-10) \times H
$
$
H=80
$
$
s=1 / 2 \times 10 \times 2^{2}
$
$
s=20 \text { and so } h=80-20=60
$
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:
Download APP for more features
1. Tons of answers.
2. Smarter Al tools enhance your learning journey.
IOS
Download
Download
Android
Download
Download
Google Play
Download
Download