question_answer

Two stones are projected from the top of a cliff h metres high, with the same speed u, so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected at an angle \[\theta \] to the horizontal, then tan \[\theta \] equals       AIEEE  Solved  Paper-2003

A)                                         \[\sqrt{\frac{2u}{gh}}\]                

B)       \[2g\,\sqrt{\frac{u}{h}}\]                             

C) \[2h\,\sqrt{\frac{u}{g}}\]                             

D) \[u\,\sqrt{\frac{2}{gh}}\]

Correct Answer: D

Solution :

When stone is projected horizontally, then                                                 \[R=u\cos 0\times t\] and        \[h=u\sin 0\times t+\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \]               \[R=ut\] and \[h=\frac{1}{2}g{{t}^{2}}\] \[\Rightarrow \]               \[R=ut\] and \[=\sqrt{\frac{2\,h}{g}}\] \[\Rightarrow \]               \[R=u\sqrt{\frac{2\,h}{g}}\]                        ?.. (i)     When stone is projected at an angle of \[\theta \] to the horizontal, then                 \[R=u\cos \theta \times t\]                         ?. (ii) and        \[h=-u\sin \theta \times t+\frac{1}{2}g{{t}^{2}}\]              ... (iii) Using Eq. (i) in Eq. (ii),we get                 \[u\sqrt{\frac{2\,h}{g}}=u\cos \theta \times t\] \[\Rightarrow \]               \[t=\frac{1}{\cos \theta }\sqrt{\frac{2\,h}{g}}\]                  ?. (iv) Using Eq. (iv) in Eq. (iii), we get                 \[h=\frac{-u\sin \theta }{\cos \theta }\sqrt{\frac{2\,h}{g}}+\frac{1}{2}g\frac{1}{{{\cos }^{2}}\theta }\left( \frac{2\,h}{g} \right)\] \[\Rightarrow \]               \[h=-u\tan \theta \sqrt{\frac{2\,h}{g}}+\frac{h}{{{\cos }^{2}}\theta }\] \[\Rightarrow \]               \[h\left( 1-\frac{1}{{{\cos }^{2}}\theta } \right)=-u\tan \theta \sqrt{\frac{2\,h}{g}}\] \[\Rightarrow \]               \[h\left( -{{\tan }^{2}}\theta  \right)=-u\tan \theta \sqrt{\frac{2\,h}{g}}\] \[\Rightarrow \]               \[\tan \theta =u\sqrt{\frac{2\,}{gh}}\]