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?

1 Answer
Jul 19, 2017

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