# Can we represent a projectile by a quadratic equation or  quadratic inequation?

### 1 Answer |Add Yours

thilina-g | College Teacher | (Level 1) Educator

Posted on

Yes we can.

Suppose an objected is projected from ground level in a velocity `V` and at angle `alpha` with the horizontal.

Therefore at t=0,

Velocity in x-direction `= V_x = Vcos(alpha)`

Velocity in y-direction `= V_y = Vsin(alpha)`

Applying,

`S =ut+1/2at^2` in x-direction.

There is no force acting on x-direction if we neglect air friction.

`x = Vcos(alpha)t+0`

`t = x/(Vcos(alpha))`

Applying,

`S =ut+1/2at^2` in y-direction.

`y = Vsin(alpha)t - 1/2gt^2`

Substituting for t,

`y = Vsin(alpha) xx x/(Vcos(alpha)) - 1/2g xx (x/(Vcos(alpha)))^2`

`y = xtan(alpha)-(gx^2)/(2V^2cos^2(alpha))`

This can be rearranged as,

`y = -(gx^2sec^2(alpha))/(2V^2)+xtan(alpha)`

sec^2(alpha) = 1+tan^2(alpha)

Therefore,

`y = -(g(1+tan^2(alpha)))/(2V^2) x^2+tan(alpha)x`

Now this is a quadratic expression of x.

We’ve answered 317,824 questions. We can answer yours, too.