Question

projectile fired at an angle equally inclined to the horizontal and vertical velocity u the horizantal range is

Answer 1

as the velocity is equally inclined to the horizontal and the vertical then the angle of projection comes out to be 45° .And the range corresponding to this angle is maximum. hence,
Range=(u^2)/g.

Answer 2

The horizontal range is equal to $R=\dfrac{u^2}{g}$.

Explanation:

The formula of the horizontal range is given by the formula as :

$R=\dfrac{u^2sin2\theta}{g}$

If the projectile fired at an angle equally inclined to the horizontal and vertical, then $\theta=45^{\circ}$

Then,

$R=\dfrac{u^2sin2(45)}{g}$

$R=\dfrac{u^2sin90}{g}$

$R=\dfrac{u^2}{g}$

So, the horizontal range is equal to $R=\dfrac{u^2}{g}$. Hence, this is the required solution.

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