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 $80 m$ in $2 s$. The distance of upper point A from the starting point is $m\left(\right.$ use $g=10 ms^{-2}$ )
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Answer: (45)
Solution:
From $A \rightarrow B$
$$ \begin{aligned} & -80=-v _1 t-\frac{1}{2} \times 10 t^{2} \\ & -80=-2 v _1-\frac{1}{2} \times 10 \times 2^{2} \\ & -80=-2 v _1-20 \\ & -60=-2 v _1 \\ & v _1=30 m / s \end{aligned} $$
From $O$ to $A$
$v^{2}=u^{2}+2 gS$
$30^{2}=0+2 \times(-10)(-S)$
$900=20 S$
$S=45 m$
