Motion in a Plane — Physics STD 11 Science — Question
Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane2 Marks
Question
Prove the following statement, “For Elevation which exceed or fall short of 45° by equal amount, the range is equal.”
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Answer
At 45°, the projectile has maximum range. At $(45^\circ-\theta)$ the range is, $\text{R}_1=\frac{\text{u}^2\sin[2(45^\circ-\theta)]}{\text{g}}$ $=\frac{\text{u}^2\sin(90^\circ-2\theta)}{\text{g}}=\frac{\text{u}^2\cos2\theta}{\text{g}}$ At $(45^\circ+\theta)$ the range is, $\text{R}_2=\frac{\text{u}^2\sin[2(45^\circ+\theta)]}{\text{g}}$ $=\frac{\text{u}^2\sin(90^\circ+2\theta)}{\text{g}}=\frac{\text{u}^2\cos2\theta}{\text{g}}$ $\text{R}_1=\text{R}_2$
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