A) their heights will be equal
B) their ranges will be equal
C) their time of flights will be equal
D) their ranges will be different
Correct Answer: B
Solution :
| Maximum height |
| \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] |
| \[\Rightarrow \] \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{\sin }^{2}}{{\theta }_{1}}}{{{\sin }^{2}}{{\theta }_{2}}}\] |
| \[=\frac{{{\sin }^{2}}{{30}^{o}}}{{{\sin }^{2}}{{60}^{o}}}\] |
| \[=\frac{{{\left( \frac{1}{2} \right)}^{2}}}{{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}=\frac{1}{3}\] |
| Range \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] |
| \[\Rightarrow \] \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{\sin \,(2\times {{30}^{o}})}{\sin (2\times {{60}^{o}})}\] |
| \[=\frac{\sin {{60}^{o}}}{\sin {{120}^{o}}}=\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}=1\] |
| \[\Rightarrow \] \[{{R}_{1}}={{R}_{2}}\] |
| Time of flight |
| \[T=\frac{2u\sin \theta }{g}\] |
| \[\Rightarrow \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\] |
| \[=\frac{\sin {{30}^{o}}}{\sin {{60}^{o}}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}\] |
| Hence, their horizontal ranges will be equal. |
| Alternative: Key Idea: For complementary angles of projection, horizontal range will be same. |
| In this problem, it is given that two particles are projected at angles \[{{30}^{o}}\] and \[{{60}^{o}}\] which are complementary angles. We know that horizontal range will be same for complementary angles. Hence, their ranges will be equal. |
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