Upon release the ball’s total energy TE(V) = 1/2 mV^2 = KE(V) the kinetic energy; where m = .4 kg and V = 12 mps. That energy is conserved at the highest point (H) where total energy TE(H) = mgH + 1/2 m (V cos(theta))^2 = PE(H) + KE(H), which is the potential energy and kinetic energy at the highest point. vx = V cos(theta), where theta = 33 degrees; this is the x component of the launch velocity. We assume vx = constant because no air drag is considered.

From the conservation of energy TE(V) = 1/2 mV^2 = mgH + 1/2 m(V cos(theta))^2 = TE(H); so the conservation of energy does in fact apply even though there is no vertical velocity. The good news is that you need not invoke it for answering a) because vx = V cos(theta) is all you need. But you do need it for b).

Solving the conservation of energy relationships we have V^2 = 2gH + V^2 cos(theta)^2; so that H = V^2(1 - cos(theta)^2)/2g = V^2 sin(theta)^2/2g H is measure above the launch point, if you wish the H to be measured from ground level you need to add in the height of the launch point above ground.

To determine its maximum height, the formula to use is

Kinetic energy = Potential energy

Kinetic energy = (1/2)mV^2

where

m = mass of the ball
V = vertical component of the initial velocity = 12sin33

Potential energy = mgh

where

m = already defined above
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
h = maximum height attained by the ball

Since KE =PE,

(1/2)(m)(12sin33)^2 = m(9.8)(h)

NOTE that since m appears on both sides of the equation, it will simply cancel out and solving for h

h = 4.9 (12sin33)^2

I trust that you can proceed with the required mathematical process from here.