The velocity as the hamper hits the water has 2 components or pieces, its velocity in the x direction and its velocity in the y direction. In the x-direction it was 107 m/s at time t=0 and it will be 107 m/s all the way down until it hits the water (there is no force in the x direction so it doesn't change). For the velocity in the y direction, it acts just like an object in free fall that is dropped from a height of 3620 meters. At t=0 its y-direction velocity is 0. Then it picks up speed as it goes down due to the force of gravity. The associated kinematic equation is vf2 = vi2 + 2ay(yf - yi) where vf is the final velocity, vi is the initial velocity, ay is the acceleration in the y direction (or g) and (yf - yi) is the change in y distance (3620 meters)
So vf2 = 02 + 2(9.8)(3620) = 70952
Taking the square root gives vf = 266.37 m/s
Now we have the horizontal velocity as 107 m/s ad the vertical velocity as 266.37 m/s like this:
To find the magnitude of the final overall velocity we use the Pythagorean Theorem to solve for the hypotenuse of the triangle (the red dotted line) which is the resultant vector.
Vel = square root (1072 + 266.372) = 287.06 m/s
Since we used 2 sig figs in g, we should really round to two sig figs but your teacher may wish you to consider 9.8 as a standard that doesn't count in which case you would use 3 sig figs from the other two givens to make the magnitude of the final vel = 287 m/s