Projectile Motion marble Problem

[LoveandHate]

Homework Statement

A marble rolls off a table with a velocity of 1.93 m/s [horiz]. The tabletop is 76.5 cm above the floor. If air resistance is negligible, determine the velocity at impact.
-\Deltat has been found in a previous question; =0.395 s
-I know the answer is 4.33 m/s [63.5 degrees below the horizontal]

Homework Equations

\Deltad=-.5a(\Deltat)²-v₂(\Deltat)

The Attempt at a Solution

\Deltad=-.5a(\Deltat)²-v₂(\Deltat)
0.765 = -.5(9.8)(.395)+v₂(.395)
but I get velocity to equal 1. something m/s?

 

[TwoTruths]

Do you want to exclusively use kinematics, or do you just want the answer? If you just want the answer, I say go the easy route: energy. I've said this before on this forum already.

Be sure to only consider the y-axis energy.

 

[LoveandHate]

i have to do it the kinematics way..
don't confuse me with the energy just yet! haha.

 

[TwoTruths]

Using kinematics makes it more confusing actually. =)

But to answer your question, if you know the time it takes to fall, and you know what acceleration it goes through, it's a simple matter of v_f(t) = v_0 + a t. Plug in initial velocity in y direction and time, then you have the y component of the velocity vector. Air resistance is nonexistent, so the horizontal velocity stays the same. Take the magnitude of this vector.

I plugged in the numbers myself (using WolframAlpha!) and got the right answer.

 

[Jebus_Chris]

\Deltad=-.5a(\Deltat)²-v₂(\Deltat)

The Attempt at a Solution

\Deltad=-.5a(\Deltat)²-v₂(\Deltat)
0.765 = -.5(9.8)(.395)+v₂(.395)
but I get velocity to equal 1. something m/s?

V2 is not final velocity, it's initial velocity and in the y direction it is equal to?