question_answer

A mountaineer standing on the edge of a cliff \[441\,\,m\]above the ground throws a stone horizontally with an initial speed of \[20\text{ }m/s\]. What is the speed with which the stone reaches the ground?

A)  \[90\text{ }m/s\]            

B)  \[95.08\text{ }m/s\]

C)  \[85\text{ }m/s\]            

D)  \[92\text{ }m/s\]

Correct Answer: B

Solution :

\[\Rightarrow \] Here \[{{U}_{x}}\] will remain constant. \[\Rightarrow \] \[V_{y}^{2}={{U}_{y}}^{2}+2{{a}_{y}}{{S}_{y}}\] \[V_{y}^{2}=0+2\times 10\times 441\] \[{{V}_{y}}=\sqrt{8820}\] \[{{V}_{y}}=94.25\,m/s\] So, let  \[U=\sqrt{V_{x}^{2}+V_{y}^{2}}\] \[=\sqrt{{{(20)}^{2}}+{{(94.25)}^{2}}}\] \[=\sqrt{400+8820}=\sqrt{9220}\] \[=96\,\,m/s\]