A) \[\frac{{{h}_{1}}t_{2}^{2}-{{h}_{2}}_{1}^{2}}{{{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}}}\]
B) \[\frac{{{h}_{1}}t_{1}^{2}-{{h}_{2}}_{2}^{2}}{{{h}_{2}}{{t}_{1}}-{{h}_{1}}{{t}_{2}}}\]
C) \[\frac{{{h}_{1}}t_{2}^{2}-{{h}_{2}}_{1}^{2}}{{{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}}}\]
D) \[\frac{{{h}_{1}}t_{1}^{2}-{{h}_{2}}_{2}^{2}}{{{h}_{1}}{{t}_{1}}-{{h}_{2}}{{t}_{2}}}\]
Correct Answer: A
Solution :
For vertically moment (for \[{{h}_{1}}\]) \[{{h}_{1}}=u\sin \theta {{t}_{1}}-\frac{1}{2}gt_{1}^{2}\] or \[{{t}_{1}}=\frac{{{h}_{1}}+\frac{1}{2}gt_{1}^{2}}{u\sin \theta }\] ?(i) \[{{h}_{2}}=u\sin \theta {{t}_{2}}-\frac{1}{2}gt_{2}^{2}\] (for \[{{h}_{2}}\]) or \[{{t}_{2}}=\frac{{{h}_{2}}+\frac{1}{2}gt_{2}^{2}}{u\sin \theta }\] ?(ii) Divide Eq. (i) by Eq. (ii) \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{{{h}_{1}}+\frac{1}{2}gt_{1}^{2}/u\sin \theta }{{{h}_{2}}+\frac{1}{2}gt_{2}^{2}/u\sin \theta }\] \[\Rightarrow \]\[{{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}}=\frac{g}{2}({{t}_{1}}t_{2}^{2}-t_{1}^{2}{{t}_{2}})\] The time of flight of the ball \[T=\frac{2u\sin \theta }{g}=\frac{2}{g}(u\,\sin \theta )\][from Eq. (i)] \[=\frac{2}{g}\left[ \frac{{{h}_{1}}+\frac{1}{2}gt_{1}^{2}}{{{t}_{1}}} \right]=\frac{2}{{{t}_{1}}}\left[ \frac{{{h}_{1}}}{g}+\frac{t_{1}^{2}}{2} \right]\] \[=\frac{{{h}_{1}}}{{{t}_{1}}}\times \frac{2}{g}+{{t}_{1}}=\frac{{{h}_{1}}}{{{t}_{1}}}\times \left( \frac{{{t}_{1}}t_{2}^{2}-t_{1}^{2}{{t}_{2}}}{{{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}}} \right)+{{t}_{1}}\] \[=\frac{{{h}_{1}}{{t}_{1}}t_{2}^{2}-{{h}_{1}}t_{1}^{2}{{t}_{2}}+{{h}_{1}}t_{1}^{2}{{t}_{2}}-{{h}_{2}}t_{1}^{3}}{{{t}_{1}}({{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}})}\] \[=\left( \frac{{{h}_{1}}t_{2}^{2}-{{h}_{2}}t_{1}^{2}}{{{h}_{1}}{{t}_{2}}-{{h}_{2}}{{t}_{1}}} \right)\]
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