A) 200 m/s
B) 100 m/s
C) 400 m/s
D) 300 m/s
Correct Answer: C
Solution :
We know that in projection of the particle, for maximum range, \[\theta =45{}^\circ \] Now maximum range \[R=\frac{{{u}^{2}}\sin 2\theta }{g}=\frac{{{u}^{2}}\sin 90{}^\circ }{g}\] \[u=\frac{g}{\sqrt{Rg}}\] ?.(1) Here, \[{{R}_{\max }}=16km=16\times {{10}^{3}}m,g=10m/{{s}^{2}}\] Now from Eq. (1), we get \[u=\sqrt{16\times {{10}^{3}}\times 10}=400\,m/s\]
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