Question

The equation of a projectile is y=root3 x -gx2/2,the angle of orojection is

Answer 1

the equation of a projectile is $\bf{y=\sqrt{3}x - \frac{gx^2}{2}}$ .

we know, if a projectile is projected with speed u at an angle $\theta$ with horizontal then, equation of trajectory of projectile is given by $\bf{y=tan\theta.x-\frac{gx^2}{2u^2cos^2\theta}}$
compare both equations ,
$\bf{tan\theta=\sqrt{3}=tan60^{\circ}}\\\bf{\theta=60^{\circ}}$

hence, angle of projection is 60°.

Answer 2

HELLO DEAR,

your questions ----------> the equation of projectile is y = √3x - gx²/2

now,

we know:-
Trajectory of projectile motion when u is intial speed inclined Ф angle with horizontal
the equation of projectile is : y = xTanФ - gx²/ 2u²Cos²Ф

and the given equation is : y = √3x - gx²/2

on comparing both the equations ,we get

TanФ = √3

so, Ф = 60°
HENCE, the angle of projectile is 60°

and 2u²Cos²Ф = 2

so u²Cos²Ф = 1

u²cos²60° = 1

u²*(1/2)² = 1

u² = 2²

so, u = 2

Hence, 2 is the initial velocity

I HOPE ITS HELP YOU DEAR,

THANKS