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2.Motion in Straight Line
hard
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 ${m}$ (use ${g}=10 \ m s^{-2}$).
A$73$
B$43$
C$75$
D$45$
(JEE MAIN-2024)
Solution
From $A \rightarrow B$$ -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}$
From $\mathrm{O}$ to $\mathrm{A}$
$ \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{gS} $
$ 30^2=0+2 \times(-10)(-\mathrm{S}) $
$ 900=20 \mathrm{~S} $
$ \mathrm{~S}=45 \mathrm{~m}$
Standard 11
Physics
