A) 50 m/s
B) 200 m/s
C) 400 m/s
D) 100 m/s
Correct Answer: C
Solution :
We know that for maximum range the value of \[\theta ={{45}^{o}}\] So, maximum range \[{{R}_{\max }}=\frac{{{u}^{2}}\sin 2\theta }{g}\] \[{{R}_{\max }}=\frac{{{u}^{2}}\sin {{90}^{o}}}{g}\] (Here, \[{{R}_{\max }}=16km=16\times {{10}^{3}}m,\]\[g=10m/{{s}^{2}}\]) or \[u=\sqrt{{{R}_{\max }}g}\] \[=\sqrt{16\times {{10}^{3}}\times 10}\] \[=400m/s\]You need to login to perform this action.
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