A projectile of mass 0.429 kg is shot from
a cannon, at height 6.6 m, as shown in the
figure, with an initial velocity vi having a
horizontal component of 6.7 m/s.
The projectile rises to a maximum height of
y above the end of the cannon’s barrel and
strikes the ground a horizontal distance x
past the end of the cannon’s barrel.
The acceleration of gravity is 9.8 m/s2 .

Determine the vertical component of the
initial velocity at the end of the cannon’s bar-
rel, where the projectile begins its trajectory.
Answer in units of m/s.

Determine the maximum height y the projectile achieves after leaving the end of the
cannon’s barrel.
Answer in units of m.

Find the magnitude of the velocity vector
when the projectile hits the ground.
Answer in units of m/s.

Use the formula v^2 = u^2+2as

You know v, the final velocity when it reaches its maximum height, the acceleration due to gravity (-9.8, negative because it is in the opposite direction of the initial velocity) and you know s, the maximum height. Solve for u, the initial velocity. It will be in terms of y.

There is not enough information. Perhaps, in the figure, more information were given. We cannot determine the value of the maximum height y only with the mass of the projectile and its horizontal component. There has to be at least the momentum of the projectile to be able to solve everything.