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