A) 5 m
B) 2.5 m
C) 1.5 m
D) 0.5 m
Correct Answer: B
Solution :
[b]: At the time of projection kinetic energy of the stone, \[K=\frac{1}{2}m{{u}^{2}}\]where m is the mass of the stone and u is the velocity of the projection. or\[{{u}^{2}}=\frac{2K}{m}=\frac{2\times 98}{2}=98\] Using,\[{{v}^{2}}={{u}^{2}}-2gh\] \[\therefore \]\[h=\frac{{{u}^{2}}}{2g}\](\[\because \]v = 0) ...(i) \[h=\frac{98}{2\times 9.8}=5m\] Also,\[K=\frac{1}{2}m{{\left( \sqrt{2gh} \right)}^{2}}\] (Using (i)) \[K'=\frac{1}{2}mv{{'}^{2}}=\frac{1}{2}m\times {{\left( \sqrt{2gh'} \right)}^{2}}\] \[\therefore \]\[\frac{\Kappa '}{K}=\frac{h'}{h}\] According to the problem \[K'=\frac{K}{2}\] \[\frac{K}{2K}=\frac{h'}{h}\] \[h'=\frac{h}{2}=\frac{5}{2}m=2.5m\]
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