Final answer:
Using the kinematic equation for constant acceleration, a rock thrown up with an initial speed of 5.0 m/s will reach a maximum height of about 1.28 meters before falling back down.
Step-by-step explanation:
To determine how high a rock will go when thrown up into the air at a speed of 5.0 m/s, we can use the principles of kinematics under the assumption of constant acceleration due to gravity (which is approximately 9.81 m/s2 directed downward). When the rock reaches its highest point, its velocity will momentarily be zero before it starts to fall back down.
Using the kinematic equation, v2 = u2 + 2as, where 'v' is the final velocity (0 m/s at the highest point), 'u' is the initial velocity (5.0 m/s), 'a' is the acceleration due to gravity (-9.81 m/s2, negative because it's opposite to the direction of the initial throw), and 's' is the displacement (height achieved), we can solve for the height 's'.
Plugging in the values, we get 0 = (5.0 m/s)2 + 2(-9.81 m/s2)s. Simplifying, we find s = (5.0 m/s)2 / (2 * 9.81 m/s2), which gives us the maximum height the rock reaches.
Calculating the exact value, s = 25.0 m2/s2 / 19.62 m/s2 = 1.276 meters. Therefore, the rock will reach a maximum height of approximately 1.28 meters from the point at which it is released.