In the case of a projectile fired at an angle equally inclined to the horizontal and vertical with velocity $u$ , the horizontal range is (Given acceleration due to gravity $= g$ )

\frac{u^{2}}{g}

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\frac{u^{2}}{2 g}

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\frac{u^{2}}{3 g}

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\frac{u^{2}}{4 g}

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Solution

The correct option is D $\frac{u^{2}}{g}$
Horizontal Range is given by,
$R = \frac{v^{2} sin 2 θ}{g}$
where $θ = 45^{\circ}$ $⟹ ∴ R = \frac{u^{2}}{g}$

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In the case of a projectile fired at an angle equally inclined to the horizontal and vertical with velocity u, the horizontal range is (Given acceleration due to gravity =g)

The correct option is D u2gHorizontal Range is given by,R=v2sin2θgwhere θ=45∘ ⟹∴R=u2g

In the case of a projectile fired at an angle equally inclined to the horizontal and vertical with velocity $u$ , the horizontal range is (Given acceleration due to gravity $= g$ )

The correct option is D $\frac{u^{2}}{g}$
Horizontal Range is given by,
$R = \frac{v^{2} sin 2 θ}{g}$
where $θ = 45^{\circ}$ $⟹ ∴ R = \frac{u^{2}}{g}$