Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
Two particles are projected from the same point with the same speed at different angles \[{{\theta }_{1}}\]and\[{{\theta }_{2}}\]to horizontal. Their times of flight are\[{{t}_{1}}\]and\[{{t}_{2}}\]and they have same horizontal range. Then (1) \[\frac{{{t}_{1}}}{{{t}_{2}}}=\tan {{\theta }_{1}}\] (2) \[{{\theta }_{1}}+{{\theta }_{2}}={{90}^{o}}\] (3) \[\frac{{{t}_{1}}}{\sin {{\theta }_{1}}}=\frac{{{t}^{2}}}{\sin {{\theta }_{2}}}\] (4) \[\frac{{{t}_{1}}}{{{t}_{2}}}=\tan {{\theta }_{2}}\]A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: A
Solution :
\[{{t}_{1}}=\frac{2u\sin {{\theta }_{1}}}{g}\] And \[{{t}_{2}}=\frac{2u\sin {{\theta }_{2}}}{g}\] As the horizontal range is same so\[{{\theta }_{1}}+{{\theta }_{2}}={{90}^{o}}\] Hence, \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\] \[=\frac{\sin {{\theta }_{1}}}{\cos ({{90}^{o}}-{{\theta }_{2}})}=\frac{\sin {{\theta }_{1}}}{\cos {{\theta }_{1}}}\] \[=\tan {{\theta }_{1}}\] Also \[\frac{{{t}_{1}}}{\sin {{\theta }_{1}}}=\frac{{{t}_{2}}}{\sin {{\theta }_{2}}}\]
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