Finding initial velocity of a cat jumping through a hoop

[Brianna I]

Homework Statement

Kit the cat is going to jump through a hoop. He begins on a wicker cabinet at a height of 1.745 m above the floor and jumps through the center of a vertical hoop, reaching a peak height 3.115 m above the floor.

With what initial velocity did Kit leave the cabinet if the hoop is at a horizontal distance of 1.567 m from the cabinet? (Express your answer in vector form)

Homework Equations

X (or Y) = v_it + 1/2at²
V_f² = V_i² + 2a(x₂-x₁)
And possibly quadratic equation.

The Attempt at a Solution

I was able to calculate the initial velocity of the y direction using V_f² = V_i² + 2a(x₂-x₁) (I know because the cat reaches peak height at 0, the vertical velocity at that time was 0).From there, I tried to solve for t. I know once I get t, I can find the initial velocity of x, but I came to an error in what I believe is my t. I tried plugging into Y = v_it + 1/2at² and made it into the quadratic formula, but got two positive answers 0.54739930925614 s and 0.51076395604999 s. Then I tried the Vf = Vi + at and got .529 s.
What can I do to find time, correctly?

 

[gneill]

I tried plugging into Y = v_it + 1/2at² and made it into the quadratic formula, but got two positive answers 0.54739930925614 s and 0.51076395604999 s. Then I tried the Vf = Vi + at and got .529 s.
What can I do to find time, correctly?

Your final method works fine. t = 0.529 s looks good.

Your other method would have worked, too if you had included the initial height. That is, the more complete version of the kinematic equation you used is:

$y(t) = y_o + v_o t + \frac{1}{2} a t^2$

Note the $y_o$ term representing the initial offset ("launch height" in this case).