Original post by EimmanuelAs a few already mention that the vertical velocity does not affect the horizontal velocity during the travel of the ball, but where the ball is located at each time, is influenced by the vertical velocity and horizontal velocity.
Allow me to take a different approach, assume the initial horizontal velocity is 5 m/s. Take the height of the tower to be 50 m and the acceleration due to gravity to be 10 m/s^2.
Initial coordinate is (0, 50)
At t = 1 s, the ball is at coordinate (5, 50 - 0.5*10*1^2) = (5, 45)
At t = 2 s, the ball is at coordinate (10, 50 - 0.5*10*2^2) = (10, 30)
At t = 3 s, the ball is at coordinate (15, 50 - 0.5*10*3^2) = (15, 5)
Next, consider a ball drop from the same tower from rest.
Initial coordinate is (0, 50)
At t = 1 s, the ball is at coordinate (0, 50 - 0.5*10*1^2) = (0, 45)
At t = 2 s, the ball is at coordinate (0, 50 - 0.5*10*2^2) = (0, 30)
At t = 3 s, the ball is at coordinate (0, 50 - 0.5*10*3^2) = (0, 5)
In both cases, the ball falls 45 m in 3 seconds in regardless of the horizontal velocity.
Hope it helps.