A) \[3.0km\]
B) \[1.5km\]
C) \[6.0km\]
D) \[0.75km\]
Correct Answer: A
Solution :
The horizontal range\[{{R}_{x}}=\frac{{{u}^{2}}\sin 2\theta }{g}\] When projected at an angle of\[{{15}^{o}}\] \[{{R}_{{{x}_{1}}}}=\frac{{{u}^{2}}}{g}\sin (2\times 15)=\frac{{{u}^{2}}}{2g}\] ... (i) When projected at an angle of\[{{45}^{o}}\] \[{{R}_{{{x}_{2}}}}=\frac{{{u}^{2}}}{g}\sin (2\times {{45}^{o}})=\frac{{{u}^{2}}}{g}\] ... (ii) According to question\[\frac{{{u}^{2}}}{2g}=1.5km\] Therefore,\[{{R}_{{{x}_{2}}}}\]which is equal to\[\frac{{{u}^{2}}}{g}\]must be\[3.0\text{ }km\].
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