Finding Speed and Angle of Projectile Trajectory

[darklich21]

Homework Statement

A diver leaves a 3m board on a trajectory that takes her 3.0m above the board, and then into the water a horizontal distance of 3.0m from the end of the board. At what speed and angle did she leave the board?

Homework Equations

Vf^2=Vi^2 + 2ad
Y=Y0 +VyT - 0.5gt^2
x=V0Cos(theta)T
Y=V0Sin(theta)T - 4.9t^2

The Attempt at a Solution

Alright, so I was able to get the correct angle, but I got the speed wrong. I got the angle by first applying Vf^2= Vi^2 + 2ad, I found Vi=7.667 m/s, plugged it into Y=Y0 + VyT -4.9t^2, got a quadratic equation, got the positive time of t= 1.889seconds, plugged this time into x=V0Cos(theta)T, and i got the angle of 78.04, this was correct.

But how Do i find the speed from the take off? Can someone help, with an explanation and an equation please?

 

[TwoTruths]

Easy way: energy.

Take the board's height as 0. At her highest point, she has no kinetic energy and all potential. At her lowest (the board), she has all kinetic and no potential.

Note that we can do this because terms like potential energy as all relative. We simply take the potential energy with respect to our newly defined equilibrium.

If you need me to, I can give you the formulas explicitly.

 

[darklich21]

Can you show me in terms of projectile motion equations and the kinematics equations?