a projectile is projected at an angle theta with the horizontal,if T is the time of flight and R is the horizontal range,then prove
T=sqrt(2Rtantheta/g) tantheta=gT^2/2R

Dear Student,

Please find below the solution to the asked query:

The equations for the time period and range are,

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

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards