A) 1 : 1
B) 2 : 3
C) 1 : 2
D) 2 : 1
Correct Answer: A
Solution :
| Key Idea: For complementary angles of projection, their horizontal ranges will be some. We know that, horizontal ranges for complementary angles of projection will be same. |
| The projectiles are projected at angles \[({{45}^{o}}-\theta )\]and \[({{45}^{o}}+\theta )\] which are complementary to each other i.e., two angles add up to give \[{{90}^{o}}\]. Hence, horizontal ranges will be equal. Thus, the required ratio is 1 : 1. |
| Alternative: Horizontal range of projectile = Horizontal component of velocity \[({{u}_{x}})\times \] Time of flight \[(T)\] |
| \[R=u\,\,\cos \,\,\alpha \times \frac{2u\,\,\sin \alpha }{g}\] |
| or \[R=\frac{{{u}^{2}}\,\,\sin 2\alpha }{g}\] |
| For \[\alpha =({{45}^{o}}+\theta ),\,{{R}_{2}}=\frac{{{u}^{2}}\sin \,2({{45}^{o}}+\theta )}{g}\] |
| \[=\frac{{{u}^{2}}\,\sin \,\,({{90}^{o}}-2\theta )}{g}\] |
| \[=\frac{{{u}^{2}}\,\cos \,2\theta }{g}\] |
| For \[\alpha =({{45}^{o}}+\theta ),\,{{R}_{2}}=\frac{{{u}^{2}}\,\sin \,2({{45}^{o}}+\theta )}{g}\] |
| \[=\frac{{{u}^{2}}\,\,\sin \,({{90}^{o}}+2\theta )}{g}\] |
| \[=\frac{{{u}^{2}}\,\cos \,\,2\theta }{g}\] |
| Hence, \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{1}\] |
| or \[{{R}_{1}}:{{R}_{2}}=1:1\] |
| Note: In the alternative method, we used the identity \[\sin ({{90}^{o}}-\theta )=\cos \theta \] and\[\sin ({{90}^{o}}+\theta )=\cos \theta \]. |
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