Q. A mountaineer standing on the edge of a cliff $ 441\,\,m $ above the ground throws a stone horizontally with an initial speed of $ 20\,m/s^{-1} $ . What is the speed with which the stone reaches the ground?
J & K CETJ & K CET 2014Motion in a Plane
Solution:
$\Rightarrow $ Here $U_{x}$ will remain constant.
$\Rightarrow V_{y}^{2}=U_{y}{ }^{2}+2 a_{y} S_{y}$
$V_{y}^{2}=0+2 \times 10 \times 441$
$V_{y}=\sqrt{8820}$
$V_{y}=94.25\, m / s ^{-1}$
So, let $U=\sqrt{V_{x}^{2}+V_{y}^{2}}$
$=\sqrt{(20)^{2}+(94.25)^{2}}$
$=\sqrt{400+8820}=\sqrt{9220}$
$=96 \,ms ^{-1}$
