The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.
Given the data in the question;
Since the brick was initially at rest before it was dropped,
Final speed of brick as it hits the ground; [tex]v = \ ?[/tex]
velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:
[tex]v^2 = u^2 + 2gh[/tex]
Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.
To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.
[tex]v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s[/tex]
Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.
Learn more about equations of motion: https://brainly.com/question/18486505
