The range of projectile thrown at an angle theta with the ground is R. The time of its flight is T.then tan theta is

Dear Student,

Please find below the solution to the asked query:

The equations for the time period and range are,

$$\begin{gathered} T=\frac{2u \sin \theta }{g} and R=\frac{u^2 \sin 2\theta }{g} \\ So from the range equation, \\ {u}^2=\frac{Rg}{\sin 2\theta }\Rightarrow u=\sqrt{\frac{Rg}{\sin 2\theta }} \\ Now, \\ T=\frac{2 sin \theta }{g}\sqrt{\frac{Rg}{sin 2\theta }}=\sqrt{\frac{4 si{n}^2 \theta }{g^2}\times \frac{Rg}{sin 2\theta }} \\ \Rightarrow T=\sqrt{\frac{2 sin \theta }{g}\times \frac{R}{cos \theta }}\Rightarrow T=\sqrt{\frac{2R \tan \theta }{g}} \\ \Rightarrow T^2=\frac{2R tan \theta }{g}\Rightarrow tan \theta =\frac{T^{2}g}{2R} \end{gathered}$$
 

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