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 \[u=\sqrt{v_{x}^{2}+v_{y}^{2}}\] \[=\sqrt{{{(20)}^{2}}+{{(94.25)}^{2}}}\] \[=\sqrt{400+8820}=\sqrt{9220}\] \[=96m/s\]You need to login to perform this action.
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