A) \[2:1\]
B) \[1:2\]
C) \[1:1\]
D) \[2:3\]
Correct Answer: C
Solution :
Horizontal range\[R=\frac{{{u}^{2}}\sin 2\theta }{g}\Rightarrow R\propto \sin 2\theta \] \[\therefore \] \[{{R}_{1}}\propto \sin 2({{45}^{o}}+\theta )\] \[{{R}_{1}}\propto \sin ({{90}^{o}}+2\,\,\theta )\] Similarly,\[{{R}_{2}}\propto \sin ({{90}^{o}}-2\,\,\theta )\] \[\because \] \[\sin ({{90}^{o}}+2\theta )=\cos 2\theta \] \[\because \] \[\sin ({{90}^{o}}-2\theta )=\cos 2\theta \] so, \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{1}\] or \[{{R}_{1}}:{{R}_{2}}=1:1\]
You need to login to perform this action.
You will be redirected in
3 sec
