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PART - 1 CH - 3 Motion in a Plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 1 CH - 3 Motion in a Plane3 Marks
Question
The time of flight of a projectile is $T$ and horizontal range is $R$. What will be the projectile angle?Or The time of flight is $T$ and horizontal range is $R$ of a projectile. Prove that the angle of projection of a projectile is $\theta=\tan ^{-1}\left(\frac{g T^2}{2 R}\right)$.
✓

Answer

Time of flight, $\quad T =\frac{2 u \sin \theta}{g}$
Horizontal range, $\quad R =\frac{u^2 \sin 2 \theta}{g}$
On square of equation (1)
$T^2=\frac{4 u^2 \sin ^2 \theta}{g}$
By dividing equation (3) to equation (2)
$\begin{array}{rlrl}
\frac{T^2}{R} & =\frac{\frac{4 u^2 \sin ^2 \theta}{g}}{\frac{u^2 \sin 2 \theta}{g}} \\
\Rightarrow \quad & \frac{g T^2}{R} & =\frac{4 \sin ^2 \theta}{\sin 2 \theta}=\frac{4 \sin ^2 \theta}{2 \sin \theta \cos \theta} \\
\Rightarrow \quad & \frac{g T^2}{R} & =\frac{2 \sin \theta}{\cos \theta}=2 \tan \theta \\
\therefore \quad & \frac{g T^2}{R} & =\tan \theta \\
\therefore \quad & \theta & =\tan ^{-1}\left(\frac{g T^2}{2 R}\right)
\end{array}$

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