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2014-09-03T04:49:21+05:30

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See diagram.

 h = 2\sqrt{gh}\ \ Sin\ \theta\ t\ -\ \frac{1}{2}\ g\ t^2 \\ \\ t^2 -
 4 \sqrt{\frac{h}{g}}\ Sin\ \theta\ t + \frac{2h}{g} = 0 \\ \\

Delta
 = \frac{16h}{g}\ Sin^2\ \theta - \frac{8h}{g} = \frac{8h}{g} [2 Sin^2\ 
\theta - 1] \\ \\ Delta = \frac{8h}{g} Cos\ (180-2\theta) \\ \\ t2\ -\ 
t1\ =\ difference\ of\ roots\ = \sqrt{Delta} = \sqrt{\frac{8h\ Cos\ 
(180-2\theta)}{g}} \\ \\

The projectile traveled 2h distance in t2 - t1  horizontally. see diagram.


 2\sqrt{gh}\ Cos\ \theta * \sqrt{\frac{8h\ Cos\ (180-2\theta)}{g}} =
 2h \\ \\ Cos\ \theta \sqrt{8 Cos(180-2\theta)} = 1 \\ \\ 8\ cos 
(180-2\theta) = Sec^2\theta \\ 8 [- cos2\theta ] = 1 + tan^2\theta \\ \\
 -8 \frac{1-tan^2\theta}{1+tan^2\theta} = 1+tan^2\theta \\ \\ 
tan^4\theta -6tan^2\theta +9 = 0 \\ \\ tan^2\theta = 3 \\ \\ tan\ theta =
 \sqrt{3} \\ \theta = 60deg, \\ \\ Substituting\ in\ t2-t1 = 
\sqrt{\frac{8h}{g}cos\ 60} = 2\sqrt{\frac{h}{g}}\\ \\


2 4 2