physics help

Don't mind the m in the pic, I was thinking forces

The distance will be half of its total travel, that's the only number you'd have. Now, what the hell is the 33o, I assume it means that the puck stays in contact with the table, but is hit 33o upwards (8o with respect to the table)

All my calculations are with respect to the slanted table, not the horizon.

Let's find the total distance

So we have
Vi = 7.922 = 8*cos(8o)
Vf = 0
It is decelerating at a rate of a = -9.8*sin(25o)
Hopefully the m will cancel out

I only use the timeless formula:
vf^2 = vi^2 + 2*a*d

0^2 = 7.922^2 + 2*-9.8*sin(25o)*(d)<solve for d

d = 7.547

We now switch to another equation mindset to find the vf halfway up the total distance.
Again, Vi = 4.357, a= -9.8*sin(25o), d = above/2, and vf = what we're solving for

vf^2 = 7.922^2+2*a*d/2


2's cancel out

vf^2 = 62.760 + (-4.142)(7.547)
vf^2 = 31.380
vfx = 5.602 m/s

My answer is in respect to the table. To find out in respect to the initial velocity you'd have to solve that all again in y terms (super bakery show fast forward)

vi=8*sin(8o)
vf = 0
a = -9.8cos(25o)
d = ?

solve for d
d = .0698

*super bakery show later*

vi = 8*sin(8o)
vf = ?
a = -9.8cos(25o)
d = .0698/2

vfy = .787

so this is all in respect to the table.

To double check my work, tan^-1( .787/5.602) is indeed 7.99, close enough to 8.