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Q. Assertion : The horizontal range is same when the angle of projection is greater than $45^{\circ}$ by certain value and less than $45^{\circ}$ by the same value.
Reason : If $\theta=45^{\circ}+\alpha$, then $R_1=\frac{u^2 \sin 2\left(45^{\circ}+\alpha\right)}{g}=\frac{u^2 \cos 2 \alpha}{g} $ $\text { If } \theta=45^{\circ}-\alpha, \text { then } R_2=\frac{u^2 \sin 2\left(45^{\circ}-\alpha\right)}{g}=\frac{u^2 \cos 2 \alpha}{g}$

Motion in a Plane

Solution:

$R =\frac{ u ^2 \sin 2 \theta}{ g } $ If $\theta=45^{\circ}+\alpha$
then $R _1=\frac{ u ^2 \sin 2\left(45^{\circ}+\alpha\right)}{ g }=\frac{ u ^2 \sin \left(90^{\circ}+\alpha\right)}{ g }$
$=\frac{ u ^2 \cos \alpha}{ g }$
If $\theta=45^{\circ}-\alpha$
then $R _2=\frac{ u ^2 \sin ^2\left(45^{\circ}-\alpha\right)}{ g }=\frac{ u ^2 \sin \left(90^{\circ}-\alpha\right)}{ g }$
$=\frac{ u ^2 \cos \alpha}{ g } \therefore R _1= R _2$