A) \[t=\frac{u-v+\sqrt{{{u}^{2}}+{{v}^{2}}}}{g}\]
B) \[t=\frac{u-v+\sqrt{{{u}^{2}}-{{v}^{2}}}}{g}\]
C) \[t=\frac{u+v+\sqrt{{{u}^{2}}-{{v}^{2}}}}{g}\]
D) \[t=\frac{u-v+\sqrt{{{u}^{2}}-{{v}^{2}}}}{2g}\]
Correct Answer: B
Solution :
| [b] Let the two bodies meet each other at a height h after time T of the projection of second body Then before meeting, the first body was in motion for time (t + T) whereas the second body was in motion for time T. |
| The distance moved by the first body in time \[\left( t+T \right)\] \[=u\left( t+T \right)-\frac{1}{2}g{{\left( t+T \right)}^{2}}\] |
| And the distance moved by the second body in time \[T=vT-\frac{1}{2}g{{T}^{2}}=h\] (supposed above) ?(1) |
| \[\therefore \] The two bodies meet each other. |
| \[\therefore \] They are equidistant from the point of projection. |
| Hence, \[u\left( t+T \right)-\frac{1}{2}g{{\left( t+T \right)}^{2}}=vT-\frac{1}{2}g{{T}^{2}}\] ... (2) |
| Also from (1) we get, \[h\,\,=\,\,vT-\frac{1}{2}g{{T}^{2}}\] |
| \[\therefore \,\,\,\frac{dh}{dT}=v-gT\] |
| \[\therefore \]h increases as T increases |
| \[\therefore \]T is minimum when h is minimum i.e., when \[\frac{dh}{dt}=0,\text{ }i.e.,\text{ when }v-gT=0\text{ or }T=v/g.\] |
| Substituting this value of T in (2), we get \[g{{t}^{2}}+2t\left( v-u \right)+2\left( v-u \right)\left( v/g \right)=0\] |
| or \[t=\frac{2g\left( v-u \right)+\sqrt{4{{g}^{2}}\left( v-u \right)+8v{{g}^{2}}\left( v-u \right)}}{2{{g}^{2}}}\] |
| or \[t=\frac{u-v+\sqrt{{{u}^{2}}-{{v}^{2}}}}{g}\] |
| Neglecting the negative sign which gives negative value of t. |
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