Intro physics course is beginning to move into 2D kinematics, stuck on the first real question.
A stone is thrown horizontally at 15m/s from the top of a cliff 44m high. What is the velocity of the rock after 2 seconds?
I tried 3(-9.8m/s2) and some other formula's but could never arrive at the correct answer of 24.7m/s. Could someone explain how to arrive at this answer and the reasoning behind it?

-- You kick it horizontally, out off of the cliff, at 15 m/s.  There's no horizontal force on it, so its horizontal speed never changes.-- As soon as it clears the edge of the cliff, gravity starts accelerating it downward, 9.8 m/s faster every second.  After 2 seconds, it has                            (2 x 9.8) = 19.6 m/s of downward vertical speed. At this point, its velocity points somewhere between horizontal and downward-vertical.  I gather from the wording of the question that you're not so much concerned with the direction of the velocity as you are with its magnitude ... that is, the stone's speed in whatever direction it's actually moving.You have a vector with two components:  15 horizontal and 19.6 vertical.Those components are perpendicular, so their resultant is just the hypotenuse of the right triangle.             (Speed)² = (15 m/s)² + (19.6 m/s)²                           = (225 m²/s²) + (384.16 m²/s²)                           =      (609.16 m²/s²) .             Speed  =  √(609.16 m²/s²)  =  24.681 m/s  That's where the 24.7 m/s comes from.    /\(I just looked back at the question again.  Notice that we don't care how high the cliff is.  That "44m" is in there only to distract those who don't know what they're doing, and who think that they must use itjust because it's there.)

ok so the x velocity is 15 m/s with no accelerationthe y velocity is 19.62 after 2 seconds because of acceleration of 9.81 m/s x 2the velocity can be found by making a diagram and solving through the pythagorean theorem15 sq + 19.62 sq = velocity sq225 + 384.9444 = velocity sqvelocity = sqrt 609.944 velocity = 24.7