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The Equation for the Path of Projectile

The horizonta...

Question

The horizontal range of a projectile is maximum when angle of projection is :

\frac{π}{3}

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\frac{π}{6}

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\frac{π}{2}

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\frac{π}{4}

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Solution

The correct option is D $\frac{π}{4}$
Range of projectile $R = \frac{u^{2} \times s i n 2 θ}{g}$
For R to be maximum $s i n 2 θ$ has to be maximum.
$s i n$ gets its maximum value at $π / 2$ ; $2 θ = π / 2 θ = π / 4$

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