Range Of Projectile

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For a given initial velocity of projection, the range is maximum when the angle of projection is:

RankGuru · Accepted Answer

1. **Range Formula**: $R = \frac{u^2 \sin(2	heta)}{g}$.
2. **Analysis**: For $R$ to be maximum, the term $\sin(2	heta)$ must be at its maximum value.
3. **Trigonometry**: The maximum value of $\sin(x)$ is $1$, which occurs when $x = 90^\circ$.
4. **Solve for $	heta$**:
   - $2	heta = 90^\circ \implies 	heta = 45^\circ$.
5. **Complementary Angles**: Note that angles $	heta$ and $90^\circ - 	heta$ give the same range, but $45^\circ$ stands unique as the peak for horizontal distance.

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Range Of Projectile

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