Physics help needed (projectile motion)

Question

A particle is projected from origin on the x-y plane with initial velocity V0 making an angle of θ with the x-axis.

If its acceleration along the x-axis is kx (k > 0) and along the y-axis it is 0, then find the velocity of the particle at any (x, y).

HINT: Neglect gravity.

Answer 1

(V0sinθ + kt , V0cosθ + 0)

Answer 2

If the acceleration is kx, then

x" = kx

or x" - k * x = 0

Solution: x(t) = A*exp(ωt) + B*exp(-ωt)

where ω = √(k)

NOTE acceleration is not constant.

x(0) = 0 means A + B = 0

x'(t) = ω*[A*exp(ωt) - B*exp(-ωt)]

x'(0) = v0*cosθ = ω*(A-B)

A = -B = v0*cosθ /2ω

x'(t) = (v0*cosθ)/2 * [exp(ωt) + exp(-ωt)]

= v0 * cosθ cosh(t*√k)

In the y direction,

y' = constant = v0*sinθ

These are your two components of velocity