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 \[h=\frac{-u\sin \theta }{\cos \theta }\sqrt{\frac{2\,h}{g}}+\frac{1}{2}g\frac{1}{{{\cos }^{2}}\theta }\left( \frac{2\,h}{g} \right)\] \[\Rightarrow \] \[h=-u\tan \theta \sqrt{\frac{2\,h}{g}}+\frac{h}{{{\cos }^{2}}\theta }\] \[\Rightarrow \] \[h\left( 1-\frac{1}{{{\cos }^{2}}\theta } \right)=-u\tan \theta \sqrt{\frac{2\,h}{g}}\] \[\Rightarrow \] \[h\left( -{{\tan }^{2}}\theta \right)=-u\tan \theta \sqrt{\frac{2\,h}{g}}\] \[\Rightarrow \] \[\tan \theta =u\sqrt{\frac{2\,}{gh}}\]You need to login to perform this action.
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