A man runs across the roof-top of a tall building and jumps horizontally with hope of landing on the roof fo the bext building which is at a lower height than the first. If his speed is ` 9 m//s`,the (horizontal ) destanc hetween the two buildings is ` 10 m` and height difference is ` 9 m`, will he be able to land on the bext building ? (taken g=10 m//s^(2)).

Explanation

Refer fig. 2 (EP) .9, let the fall of man from ` o to A` takes ` t seconds. The vertical distanc ecoverd is ` 9 m`. Taking vertical downward motion from `O to A`, we have
` y_0 =0, y= 9 m, u_y =0, a_y =10 m//s^2, t=?`
As ` y=y_0 = u-y t + 1/2 a-y t^2`
:. ` 9 =0 +0 + 1/2 xx 10 xx t^2 =5 =5 t^2`
or ` t = sqrt 9 /5 = sqrt 3 /9sqrt 5) s`
Horizonal distance travelled in (t) seconds is
` = 9 xx 3 /(sqrt) = (27 sqrt 5)/5 ~~ 12 m gt 10 m`
Therefore, he will land on the next building .