A) \[3{{y}_{1}}\]
B) \[2{{y}_{1}}\]
C) \[\frac{{{y}_{1}}}{2}\]
D) \[\frac{{{y}_{1}}}{3}\]
Correct Answer: D
Solution :
For complementary angles ranges are equal. \[\therefore \] \[{{\theta }_{1}}=\frac{\pi }{3},\] then \[{{\theta }_{2}}=\frac{\pi }{2}-\frac{\pi }{3}=\frac{\pi }{6}\] Maximum height for first stone \[{{y}_{1}}=\frac{{{u}^{2}}{{\sin }^{2}}{{\theta }_{1}}}{2g}\] \[{{y}_{1}}=\frac{{{u}^{2}}{{\sin }^{2}}\frac{\pi }{3}}{2g}\] \[=\frac{{{u}^{2}}{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}}{2g}=\frac{3{{u}^{2}}}{8g}\] \[\Rightarrow \] \[{{u}^{2}}=\frac{8g\,{{y}_{1}}}{3}\] Maximum height for second stone \[{{y}_{2}}=\frac{{{u}^{2}}{{\sin }^{2}}{{\theta }_{2}}}{2g}\] \[=\frac{{{u}^{2}}{{\sin }^{2}}\frac{\pi }{6}}{2g}\] \[=\frac{{{u}^{2}}{{\left( \frac{1}{2} \right)}^{2}}}{2g}=\frac{{{u}^{2}}}{8g}\] \[{{y}_{2}}=\frac{1}{8g}.\frac{8g\,{{y}_{1}}}{3}\] \[=\frac{{{y}_{1}}}{3}\]
You need to login to perform this action.
You will be redirected in
3 sec
