M2 - Jan 2014 IAL edexcel question 6a help!

I dont understand what the mark scheme or the model answers are saying for this paper. Exam solutions doesnt do it either.

6a.
Paper:
https://ca99c64778b62ba7e7b339967029e090c5733c3a.googledrive.com/host/0B1ZiqBksUHNYOXdJSDFGUUZPRUE/January%202014%20(IAL)%20QP%20-%20M2%20Edexcel.pdf

MS:
https://ca99c64778b62ba7e7b339967029e090c5733c3a.googledrive.com/host/0B1ZiqBksUHNYOXdJSDFGUUZPRUE/January%202014%20(IAL)%20MS%20-%20M2%20Edexcel.pdf

What is the kinetic energy at the beginning? It has a kinetic of 1/2 m(3^2 + v^2) at the beginning by adding the KE of the horizontal and vertical component separately.

Not use suvat, horizontally the speed stats the same throughout the motion. So horizontal component is still 3 ms^(-1).

Vertically, use v = u +at, so the new speed is v - gt since the initial speed is v and the acceleration is -g.

Now the this means the K.E at this point is 1/2 m (3^2 + (v-gt)^2).

But you know the KE at this point must be half the initial K.E. That is the initial K.E is twice the new KE.

So 1/2m (3^2 + v^2) = 2 * (1/2 m (3^2 + (v-gt)^2)

so cancelling out the 1/2m we get:

3^2 + v^2 = 2(3^2 + (v-gt)^2)

now expand, get a quadratic in v and solve.

Oh ok. Thanks