Final answer:
The rock will be traveling at a velocity of approximately 19.10 m/s just before it hits the ground. The horizontal motion is unaffected by gravity and remains at a constant velocity of 7.2 m/s. The vertical motion can be described using equations of motion in physics.
Step-by-step explanation:
The rock will be traveling at a velocity of approximately 10.3 m/s just before it hits the ground. To find this answer, we can use the equations of motion in physics. Since the rock is thrown horizontally, the horizontal motion is unaffected by gravity and remains at a constant velocity of 7.2 m/s. The vertical motion can be described by the equation: h = ut + (1/2)gt^2, where h is the height, u is the initial velocity, t is the time, and g is the acceleration due to gravity.
Using the given information, we can calculate the time it takes for the rock to hit the ground: t = √(2h/g) = √(2(15 m)/9.8 m/s^2) ≈ 1.77 s. Now, we can find the final vertical velocity of the rock just before hitting the ground using the equation: v = u + gt = 0 + (9.8 m/s^2)(1.77 s) ≈ 17.36 m/s.
Since the rock is thrown horizontally, the horizontal and vertical motions are independent. Therefore, the horizontal velocity of 7.2 m/s remains constant. The total velocity of the rock just before hitting the ground can be calculated using the Pythagorean theorem: v = √(v_h^2 + v_v^2) = √((7.2 m/s)^2 + (17.36 m/s)^2) ≈ 19.10 m/s. Therefore, the rock will be traveling at a velocity of approximately 19.10 m/s just before it hits the ground.