A) 400 m/s
B) 200 m/s
C) 100 m/s
D) 50 m/s
Correct Answer: A
Solution :
For a body moving in projectile motion, the horizontal range covered is
\[\Delta {{i}_{C}}=0.96\times 7.2=6.91mA\] \[\Delta {{i}_{E}}=\Delta {{i}_{C}}+\Delta {{i}_{B}}\] For maximum range \[\therefore \] \[\Delta {{i}_{B}}=\Delta {{i}_{E}}-\Delta {{i}_{C}}\] \[=7.2-6.91=0.29mA\] Given, \[\frac{(2n+1)\lambda }{2}\] \[\frac{(n+1)\lambda }{2}\] \[n(\lambda +1)\] \[n\lambda \] \[[M{{L}^{3}}{{T}^{-3}}]\] \[[M{{L}^{-1}}{{T}^{-1}}]\] Note: Since \[[M{{L}^{2}}{{T}^{-2}}]\] This is why a long jumper takes jump at an angle of \[[M{{T}^{-2}}]\]. Substituting \[\sqrt{\frac{1}{{{\varepsilon }_{0}}{{\mu }_{0}}}}\]in place of \[\sqrt{\frac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}}\], we have \[\frac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}\] \[{{\varepsilon }_{0}}{{\mu }_{0}}\] \[{{V}_{0}}\] From this it is clear that horizontal range is same whether body is projected at \[{{V}_{e}}\] or \[{{V}_{e}}=2{{V}_{0}}\]
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