Final answer:
The speed of a 2.0 kg dropped from 19 m is found using conservation of energy or kinematic equations, resulting in approximately 19.3 m/s just before it hits the ground.
Step-by-step explanation:
The question asks us to calculate the speed of a 2.0 kg ball just before it hits the ground after being dropped from a height of 19 meters. To find this, we can use the principles of conservation of energy or the kinematic equations for objects in free fall. Assuming no air resistance, all potential energy (PE) of the ball at the initial height will be converted to kinetic energy (KE) just before impact. The equation PE = KE can be represented as mgh = (1/2)mv^2, where m is mass, g is acceleration due to gravity (9.80 m/s2), h is height, and v is velocity. Solving for v we have v = √(2gh), substituting the known values gives us:
v = √(2 * 9.80 m/s2 * 19 m) = √(372.4 m2/s2) = 19.3 m/s.
Therefore, the speed of the ball just before impact is approximately 19.3 m/s.