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 \[R=\frac{{{u}^{2}}\sin 2\theta }{g}=\frac{{{u}^{2}}\sin 90{}^\circ }{g}\] The muzzle velocity of the shell \[u=\sqrt{R}g\] ...(i) Here, \[{{R}_{\max }}=16\text{ }km=16\times {{10}^{3}}m,\text{ }g=10\text{ }m/{{s}^{2}}\] Now, from Eq. (i), we get \[u=\sqrt{16\times {{10}^{3}}\times 10}\] \[=400\text{ }m/s\]
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