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A projectile is projected with velocity of $25 m/s$ at an angle $\theta$ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of $\theta$ will be: [use $g = 10 m/s^{2}$ ]
Explanation
At maximum height inclination with horizontal becomes. So, $t = u \sin \theta/g \Rightarrow u = gt/\sin \theta$ Now, $R = u \cos \theta \times T \Rightarrow R = u \cos \theta \times 2t$ $\Rightarrow \cos \theta = R/2ut \Rightarrow \cos \theta = R \sin \theta/(2t \cdot gt)$ [From (i)] $\Rightarrow \tan \theta = 2gt^{2}/R \Rightarrow \tan \theta = 20t^{2}/R \Rightarrow \cot \theta = R/20t^{2}$