Question

The equation of the trajectory of particle is `x=sqrt3 y-5y^2`.The particle is at t=0 at origin.Its acceleration is `veca=-10 hati`.What is its initial velocity?

Answer 1

Here given acceleration is `-10hati` i.e it is acting along negative direction of x- axis. Here equation of trajectory is quadratic in terms of y. This means we get two values of y for one value of x.

Let the velocity of projection of the particle from the origin (0,0) be `u` making an angle of projection `alpha` with the positive direction of y-axis at `t=0`

So resolved components are `u_y=ucosalpha and u_x=usinalpha`.

Let the particle takes time `t` to reach at P (x,y)

So

`x=u_x xxt-1/2xxaxxt^2`

`x=usinalpha xxt-1/2xxaxxt^2......[1]`

and

`=>y=u_y xxt=ucosalphat.....[2]`

Combining [1] and[2] we get

`x=usinalphaxxy/(ucosalpha)-1/2xxaxxy^2/(ucosalpha)^2`

`x=tanalphaxxy-1/2xx10xxsec^2alpha/u^2y^2`

`x=tanalphaxxy-5xxsec^2alpha/u^2y^2.....[3]`

So we obtain this equation of trajectory.

Again the given equation of trajectory is

`x=sqrt3y-5y^2.....[4]`

Comparing [3] and [4] we get

`tanalpha=sqrt3`

and

`5xxsec^2alpha/u^2=5`

`u^2=sec^2alpha=1+tan^2alpha=1+(sqrt3)^2=4`

So `u=2` unit

Hence initial velocity `u=2` unit