A) 16 m
B) 10 m
C) 8m
D) 18m
Correct Answer: B
Solution :
Key Idea: At maximum height vertical component of final velocity is zero. It is given velocity at half the height is 10 m/s. From equation of motion, we have \[{{v}^{2}}={{u}^{2}}-2gs\] where v is final velocity, g is acceleration and s is displacement. At maximum height v = 0 \[\therefore \] \[{{u}^{2}}=2\,g\,s\] \[\Rightarrow \] \[s=\frac{{{u}^{2}}}{2g}\] At half the height, \[s'=\frac{s}{2}=\frac{1}{2}\left( \frac{{{u}^{2}}}{2g} \right)\] Now \[100-{{u}^{2}}=2\times (-g)\times \frac{{{u}^{2}}}{4g}\] \[\Rightarrow \] \[u=\sqrt{200}\,m/s\] Maximum height attained is \[=\frac{200}{(2\times 10)}=10\,\text{m}\]
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