Check work on simple Energy problem

[Nivlac2425]

Homework Statement

A particle of mass m starts at rest and slides down a frictionless track as shown. It leaves the track horizontally, striking the ground as indicated. At what height h did it start above the ground?
Image: http://tinypic.com/view.php?pic=2irb03r&s=4

Homework Equations

The Attempt at a Solution

To find h, I started by equating the energies at the top and bottom as PE=KE, mgh=1/2mv^2, and to solve for the height, I needed to solve for the velocity, v, at the bottom.
I used a kinematic equation for the y-direction(Vf^2=Vo^2+2ad) where Vo=0, Vf= vsintheta, a=g, and d=1.25m and solved for v^2. I then plugged this v^2 into my energy equation and solved h to be 3.2m.

I am just wondering whether setting Vf=vsintheta was correct, assuming the velocity needed at the bottom is the total resultant velocity, not just the vertical component of it.
Can someone please verify this and my work and answer as well?

PS: I realize the value of theta is required for calculation and I have found that to be equal to 38.66 degrees using tan of theta

Thank you PF community!

 

[drizzle]

there are 3 different velocities, not just two;

the initial velocity V_o where the particle starts at rest, another one [say V_f1] where it leaves the track horizontally [that means the angle is…?] and the last V_f2 where it strikes the ground [note that the path this particle takes once it leaves the track is the projectile motion,‏where you treat V_f1 as an initial velocity].

 

[rl.bhat]

When the particle leaves the track horizontally, its vertical velocity is zero. Horizontal velocity is, say v. Find the time t taken by the particle to reach the ground at the depth h1. Horizontal distance traveled by the particle is given. From that find v by using the formula v = d/t. Find the height h2 through which the particle must fall to have velocity v. Finally find h = h1 + h2