Final answer:
The magnitude of the velocity of a rock thrown horizontally with an initial speed of 20 m/s after two seconds is approximately 28 m/s, considering the constant horizontal velocity and the increased vertical velocity due to gravity.
Step-by-step explanation:
When you throw a rock horizontally at a cliff with a speed of 20 m/s, after two seconds, you can calculate the magnitude of the velocity of the rock by considering the horizontal and vertical components of motion separately since they are independent. The horizontal velocity remains constant because there are no horizontal forces acting on the rock (assuming no air resistance), so it is still 20 m/s after 2 seconds. The vertical velocity can be found using the equation v = gt, where g is the acceleration due to gravity (9.8 m/s2) and t is the time in seconds. After 2 seconds, the vertical velocity is v = 9.8 m/s2 × 2 s = 19.6 m/s. To find the total velocity, you use the Pythagorean theorem because the horizontal and vertical motions are at right angles to each other.
Total velocity (v) = √(vx2 + vy2) = √(202 + 19.62)
= √(400 + 384.16) = √(784.16) = 28.0 m/s.
Therefore, the magnitude of the velocity of the rock is closest to 28 m/s, not 37 m/s as was suggested in the question.