A man runs across the roof, top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is at a lower height than the first. If his speed is $9\, m/s$. the (horizontal) distance between the two buildings is $10\, m$ and the height difference is $9\, m$, will be able to land on the next building ? $($ Take $g = 10 \,m/s^2)$
A man runs across the roof, top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is at a lower height than the first. If his speed is $9\, m/s$. the (horizontal) distance between the two buildings is $10\, m$ and the height difference is $9\, m$, will be able to land on the next building ? $($ Take $g = 10 \,m/s^2)$
Given, horizontal speed of the $\operatorname{man}\left(u_{x}\right)=9 \mathrm{~m} / \mathrm{s}$
Horizontal distance between the two buildings, $x=10 \mathrm{~m}$
Height difference between the two buildings, $y=9 \mathrm{~m}$
and $g=10 \mathrm{~m} / \mathrm{s}^{2}$
Let the man jumps from point $\mathrm{A}$ and land on the roof of the next building at point $\mathrm{B}$. Taking motion in vertical direction,
$y=u t+\frac{1}{2} a t^{2}$
$\therefore 9=0 \times t+\frac{1}{2} \times 10 \times t^{2} \quad\left(\because u=u_{v}=0\right)$
$\therefore 9=5 t^{2}$
$t=\sqrt{\frac{9}{5}}=\frac{3}{\sqrt{5}}$
$=9 \times \frac{3}{\sqrt{5}}$
$=\frac{27}{\sqrt{5}} m=12 m$
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