Motion In One Dimension Question 6
Question 6 - 2024 (27 Jan Shift 2)
A body falling under gravity covers two points $A$ and $B$ separated by $80 \mathrm{~m}$ in $2 \mathrm{~s}$. The distance of upper point A from the starting point is $\mathrm{m}\left(\right.$ use $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
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Answer: (45)
Solution:
From $\mathrm{A} \rightarrow \mathrm{B}$
$$ \begin{aligned} & -80=-\mathrm{v}{1} \mathrm{t}-\frac{1}{2} \times 10 \mathrm{t}^{2} \ & -80=-2 \mathrm{v}{1}-\frac{1}{2} \times 10 \times 2^{2} \ & -80=-2 \mathrm{v}{1}-20 \ & -60=-2 \mathrm{v}{1} \ & \mathrm{v}_{1}=30 \mathrm{~m} / \mathrm{s} \end{aligned} $$
From $\mathrm{O}$ to $\mathrm{A}$
$\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gS}$
$30^{2}=0+2 \times(-10)(-S)$
$900=20 \mathrm{~S}$
$\mathrm{S}=45 \mathrm{~m}$
