For me, first I resolved the two component vectors. The hypotenuse is 4.534, so using SOH CAH TOA I found the Y component to be 1.916m/s and the X to be 4.109m/s. Then, I used the distance formula

Dy = VoT + 1/2(g)(t^2)

and solved for T to get

1.51 = (1.916)(t) + 1/2(9.80)(x^2)

Then T = .393

Then to solve for Dx I used

Dx = VoT

the acceleration part of the equation is eliminated because their is constant uniform velocity on the X component vector.

Dx = (4.109)(.393)

Dx = 1.62m

This answer is wrong though