A) \[{{t}_{1}}+{{t}_{2}}=\frac{2R}{g}\]
B) \[{{t}_{1}},{{t}_{2}}=\frac{2R}{g}\]
C) \[{{t}_{1}}-{{t}_{2}}=\frac{2R}{g}\]
D) \[{{t}_{1}}\,{{t}_{2}}=\frac{2R}{g}\]
Correct Answer: D
Solution :
[d] Let \[\theta \] and \[90{}^\circ -~\theta \]be the angles of projection for the same range \[{{t}_{1}}=\frac{2u\,\sin \,\theta }{g}\,and\,\,{{t}_{2}}=24\,Sin\,\frac{[90{}^\circ -\theta ]}{g}=\frac{2\,u\,Cos\,\theta }{g}\] \[{{t}_{1}}{{t}_{2}}=\frac{4{{u}^{2}}\,\sin \,\theta \,Cos\,\theta }{g}\,=\frac{2R}{g}\] \[\therefore \,\,\left[ R=\frac{2{{u}^{2}}\,Sin\theta \,Cos\theta }{g} \right]\]You need to login to perform this action.
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