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 the horizontal. They have the same horizontal range. Their times of flights are\[{{t}_{1}}\]and\[{{t}_{2}}\]respectively. Then (1) \[{{\theta }_{1}}+{{\theta }_{2}}={{90}^{o}}\] (2) \[\frac{{{t}_{1}}}{\sin {{\theta }_{1}}}=\frac{{{t}_{2}}}{\sin {{\theta }_{2}}}\] (3) \[\frac{{{t}_{1}}}{{{t}_{2}}}=\tan {{\theta }_{1}}\] (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 :
Projectiles have the same horizontal range for complimentary angles of projection. \[\therefore \] \[{{\theta }_{1}}+{{\theta }_{2}}={{90}^{o}}\] \[{{\theta }_{2}}={{90}^{o}}-{{\theta }_{1}}\] The time of flight for first particle; \[{{t}_{1}}=\frac{2u\sin {{\theta }_{1}}}{g}\] The time of flight for second particle \[{{t}_{2}}=\frac{2u\sin {{\theta }_{2}}}{g}\] \[\therefore \] \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\] Or \[\frac{{{t}_{1}}}{\sin {{\theta }_{1}}}=\frac{{{t}_{2}}}{\sin {{\theta }_{2}}}\] Again, \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\] Or \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\sin {{\theta }_{1}}}{\sin (90-{{\theta }_{1}})}\] \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\sin {{\theta }_{1}}}{\cos ({{\theta }_{1}})}\] \[\therefore \] \[\frac{{{t}_{1}}}{{{t}_{2}}}=\tan {{\theta }_{1}}\] Hence, option is correct.
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