Motion in a Plane — Physics STD 11 Science — Question
Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane3 Marks
Question
Show that the horizontal range of a projectile is same for angles of projection $(45+\alpha)^\circ$ and $(45-\alpha)^\circ.$
✓
Answer
For angle of projection $(45+\alpha)^\circ,$ the horizontal range is $\text{R}_1=\frac{\text{u}^2}{\text{g}}\sin2(45+\alpha)^\circ$ $=\frac{\text{u}^2}{\text{g}}\sin(90+2\alpha)^\circ=\frac{\text{u}^2}{\text{g}}\cos2\alpha\dots(\text{i})$ and for an angle of projection
$(45-\alpha)^\circ,$ the horizontal range is $\text{R}_1=\frac{\text{u}^2}{\text{g}}\sin2(45-\alpha)^\circ$
$=\frac{\text{u}^2}{\text{g}}\sin(90-2\alpha)^\circ=\frac{\text{u}^2}{\text{g}}\cos2\alpha\dots(\text{ii})$ By comparing eqn. (i) and (ii), we find $R_1= R_2$
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