The path followed by a body projected along y-axis is given as by y = √3x -(1 / 2)x² , if g = 10 m/s, then the initial velocity of projectile will be – (x and y are in m)

The correct option is 2√(10) m/s.

Step-by-step Derivation:

The given equation of the path (trajectory) followed by the projectile is:

$y=\sqrt{3}x-\frac{1}{2}{x}^2$

The standard equation of the trajectory of a projectile projected with an initial velocity $u$ at an angle $\theta$ with the horizontal (x-axis) is given by:

$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

By comparing the coefficients of $x$ and $x^2$ in the given equation and the standard equation, we get:

1) Comparing the coefficient of $x$:
$\tan \theta =\sqrt{3}$
This gives the angle of projection:
$\theta =60°$

2) Comparing the coefficient of $x^2$:
$\frac{g}{2u^2{\cos }^2\theta }=\frac{1}{2}$

Given the acceleration due to gravity $g=10{\text{m/s}}^2$ and $\theta =60°$, we can substitute these values into the equation:

$\frac{10}{2u^2{\cos }^2(60°)}=\frac{1}{2}$

Since $\cos (60°)=\frac{1}{2}$, we have ${\cos }^2(60°)=\frac{1}{4}$. Substituting this value in:

$\frac{10}{2u^2\left(\frac{1}{4}\right)}=\frac{1}{2}$

Simplifying the denominator on the left-hand side:

$\frac{10}{\left(\frac{u^2}{2}\right)}=\frac{1}{2}$

$\frac{20}{u^2}=\frac{1}{2}$

Cross-multiplying to solve for $u^2$:

$u^2=40$

Taking the square root on both sides:

$u=\sqrt{40}=\sqrt{4\times 10}=2\sqrt{10}\text{m/s}$

Thus, the initial velocity of the projectile is 2√(10) m/s.