A block is launched up a frictionless 40 slope with an initial speed v and reaches a maximum vertical height..

A block is launched up a frictionless 40 slope with an initial speed v and reaches a maximum vertical height h. The same block is launched up a frictionless 20 slope with the same initial speed v. On this slope, the block reaches a maximum vertical height of ?

I know that they are the same, but don't understand why.

This is about energy balance. Kinetic energy gets transferred into potential energy. If both scenarios start off with the same kinetic energy, m and v, then they'll have the same potential energy in the end.

1/2*m*v^2 = m*g*h
1/2*v^2 = g*h

h = v^2 / (2*g)

Long way...
Sum of the forces in the direction of motion = m*a = -m*g*sin(theta)
a = g*sin(theta)

equation of motion
vf^2 = vi^2 + 2*a*d

d, in this case, is the distance along the ramp, which can be expressed in terms of h

sin(theta) = h / d ==> d = h/sin(theta)

In this case, the block is coming to rest, so vf = 0

0 = v^2 + 2*g*sin(theta)*h/sin(theta)
0 = v^2 + 2*g*h
h = v^2/(2*g)

So, no matter the angle is, the angles will get canceled out and you will wing up with h as a function of v.