A) They meet at time \[t=\frac{u}{g}\] and at a height \[\frac{{{u}^{2}}}{2g}+\frac{g{{T}^{2}}}{8}\]
B) They meet at time\[t=\frac{u}{g}+\frac{T}{2}\] and at a height \[\frac{{{u}^{2}}}{2g}+\frac{g{{T}^{2}}}{8}\]
C) They meet at time \[t=\frac{u}{g}+\frac{T}{2}\] and at a height \[\frac{{{u}^{2}}}{2g}-\frac{g{{T}^{2}}}{8}\]
D) They never meet
Correct Answer: C
Solution :
[c] For first projectile, \[{{h}_{1}}=ut-\frac{1}{2}g{{t}^{2}}\] For second projectile, \[{{h}_{2}}=u(t-T)-\frac{1}{2}g{{(t-T)}^{2}}\] When both meet i.e. \[{{h}_{1}}\]=\[{{h}_{2}}\]: \[ut-\frac{1}{2}g{{t}^{2}}=u(t-T)-\frac{1}{2}g{{(t-T)}^{2}}\] \[\Rightarrow uT+\frac{1}{2}g{{T}^{2}}=gtT\Rightarrow t=\frac{u}{g}+\frac{T}{2}\] And \[{{h}_{1}}=u\left( \frac{u}{g}+\frac{T}{2} \right)-\frac{1}{2}g{{\left( \frac{u}{2}+\frac{T}{2} \right)}^{2}}=\frac{{{u}^{2}}}{2g}-\frac{g{{T}^{2}}}{8}\]
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