Question

The equation of projectile is y=√3x-gx^2/2 the horizontal range is

Answer 1

HELLO DEAR,

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

now,

we know:-

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°

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

Answer 2

We will solve it like this:

y = xTanQ - gx² / 2u² Cos²Q

The equation is: y = √3x - gx²/2

So when we solve the equation we will get:

TanQ = √3

So Q = 60°

and u² Cos² Q = 1

u² Cos² 60 = 1

u² = 4

so u will be 2

So 2 is the initial velocity.

If there is any confusion please leave a comment below.