A) 2 : 1
B) 1 : 1
C) 1 : cos \[\text{ }\!\!\theta\!\!\text{ }\]
D) 1 : sec \[\text{ }\!\!\theta\!\!\text{ }\]
Correct Answer: B
Solution :
Let \[{{u}_{1}}\] and \[{{u}_{2}}\] be the initial velocities. If \[{{s}_{1}}\] and \[{{s}_{2}}\] are the heights attained by them, then \[{{s}_{1}}=\frac{{{u}_{1}}^{2}}{2g}\] and \[{{s}_{2}}=\frac{{{u}_{2}}^{2}{{\sin }^{2}}\theta }{2g}\] ... (i) Since their time of ascent are equal, hence \[t=\frac{{{u}_{1}}^{2}}{g}=\frac{{{u}_{2}}\sin \theta }{g}\] \[{{u}_{1}}={{u}_{2}}\sin \theta \] ... (ii) From equation (i),\[\frac{{{s}_{1}}}{{{s}_{2}}}=\frac{{{u}_{1}}^{2}}{{{u}_{2}}^{2}{{\sin }^{2}}\theta }\] ... (iii) From equations (ii) and (iii) we get, \[\frac{{{s}_{1}}}{{{s}_{2}}}=\frac{{{({{u}_{2}}\sin \theta )}^{2}}}{{{u}_{2}}^{2}{{\sin }^{2}}\theta }=\frac{1}{1}\]
You need to login to perform this action.
You will be redirected in
3 sec
