Question
Prove that in projectile motion, when bodies are projected at angle $\left(45^{\circ}+\phi\right)$ and $\left(45^{\circ}-\phi\right)$ from the horizontal, their horizontal range will be same.
✓
Answer
We know that by projecting an object at angle $\theta$,
$R=\frac{u^2 \sin 2 \theta}{g}$
But it is given that when object is projected at angle $(45+\phi)$ and $\left(45^{\circ}-\phi\right)$ from the horizontal, their ranges are same.$
\begin{aligned}
R_1 & =\frac{u^2 \sin 2\left(45^{\circ}+\phi\right)}{g} \\
R_1 & =\frac{u^2 \sin \left(90^{\circ}+2 \phi\right)}{g} \\
R_1 & =\frac{u^2 \cos 2 \phi}{g}......(1) \\
\text { Similarly, } \quad R_2 & =\frac{u^2 \sin 2\left(45^{\circ}-\phi\right)}{g} \\
R_2 & =\frac{u^2 \sin \left(90^{\circ}-2 \phi\right)}{g} \\
R_2 & =\frac{u^2 \cos 2 \phi}{g}.......(2)
\end{aligned}$
From equations (1) and (2), it is clear
$R_1=R_2$
$R=\frac{u^2 \sin 2 \theta}{g}$
But it is given that when object is projected at angle $(45+\phi)$ and $\left(45^{\circ}-\phi\right)$ from the horizontal, their ranges are same.$
\begin{aligned}
R_1 & =\frac{u^2 \sin 2\left(45^{\circ}+\phi\right)}{g} \\
R_1 & =\frac{u^2 \sin \left(90^{\circ}+2 \phi\right)}{g} \\
R_1 & =\frac{u^2 \cos 2 \phi}{g}......(1) \\
\text { Similarly, } \quad R_2 & =\frac{u^2 \sin 2\left(45^{\circ}-\phi\right)}{g} \\
R_2 & =\frac{u^2 \sin \left(90^{\circ}-2 \phi\right)}{g} \\
R_2 & =\frac{u^2 \cos 2 \phi}{g}.......(2)
\end{aligned}$
From equations (1) and (2), it is clear
$R_1=R_2$
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