A) \[200\,\,m/s\]
B) \[100\,\,m/s\]
C) \[400\,\,m/s\]
D) \[300\,\,m/s\]
Correct Answer: C
Solution :
Key Idea: For maximum horizontal range body should be projected at an angle of\[{{45}^{\circ }}\]. The horizontal range covered is given by \[R={{u}_{x}}\times T\] \[=u\,\,\cos \,\,\theta \times \frac{2u\,\,\sin \,\theta }{g}\] \[={{u}^{2}}\,\,\frac{\left( 2\,\sin \,\theta \,\,\cos \,\theta \right)}{g}\] \[R=\frac{{{u}^{2}}\,\sin \,20}{g}\] For\[{{R}_{\max }}\,\sin \,2\theta =1\], \[\therefore \]\[\theta ={{45}^{\circ }}\] \[\therefore \] \[{{R}_{\max }}=\frac{{{u}^{2}}}{g}\] \[\Rightarrow \] \[u=\sqrt{{{R}_{\max }}g}\] Given,\[{{R}_{\max }}=16\,km=16\times {{10}^{3}}\,m,\,g=10\,m/{{s}^{2}}\] \[u=\sqrt{16\times {{10}^{3}}\times 10}\] \[u=400\,m/s\] Note: Horizontal range is same whether the body is projected at \[\theta \] or \[\left( {{90}^{\circ }}-\theta \right)\,i.e.,\]for complementary angles of projection range is same.You need to login to perform this action.
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