question_answer

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

A)  90 m/s            

B)  95.08 m/s

C)  85 m/s            

D)  92 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 \[u=\sqrt{v_{x}^{2}+v_{y}^{2}}\] \[=\sqrt{{{(20)}^{2}}+{{(94.25)}^{2}}}\] \[=\sqrt{400+8820}=\sqrt{9220}\] \[=96m/s\]