How can the momentum of a thrown rock be calculated with given parameters?

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To calculate the momentum of a thrown rock, the mass of the rock (1.10 kg) and its launch angle (39.0°) are crucial parameters. The initial velocity was calculated using the formula V = √2gx, yielding a velocity of 22.54 m/s, leading to a momentum of 24.8 kg·m/s. However, the method used is inappropriate since it assumes the rock is dropped from rest, neglecting the effects of the launch angle and initial height. The irregular shape of the rock and its height above sea level may also influence its trajectory and momentum. Understanding these factors is essential for an accurate calculation of the rock's momentum at the moment of release.
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Homework Statement
A rock of mass 1.10kg and volume 283cm3 is irregular shape. A person throws the rock with a launch angle of 39.0° (above the horizontal) over flat level ground from a height of 0.668m above the ground. Ground level is at an elevation of 138m above sea level. The rock goes 7.40 m down range before it hits the ground. The rock bounces off the ground achieving a maximum height of 12.0cm and traveling an additional 18.0cm down range before coming to rest with a thud. Find the momentum of the rock at the instant it leaves the thrower’s hand.

The attempt at a solution
So what I did was V = √2gx (where x is distance) and I got a velocity of 22.54 m/s. I plugged that into the equation P=m*v to get momentum of 24.8 kgm/s However, I'm not sure I did this right at all.
 
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Welcome to PF.
Hint. Look up "ballistics".
the formula you used is for when the rock is dropped a from rest from a height x, when gravity is the only force. Is that what happened here?

Why is it important that the rock is irregular?
Why is the height above sea level mentioned?
What difference would the angle the rock got thrown make?
 

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