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,
- Initial Velocity;
- Height from which it has dropped;
- Gravitational field strength;
Final speed of brick as it hits the ground; 
<h3>Velocity</h3>
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:
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.
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: brainly.com/question/18486505
Answer:
Explanation: Bagong bayani ang turing sa ating mga Overseas Filipino. Workers (OFWs) sapagkat ang kanilang pagpapagod at pakikipagsapalaran sa ibayong dagat ay hindi lamang ... kasigurahan ang pagkukunan ni Lito sa panahon ng kanyang.
Answer:
Wait, that can happen? I'm sorry.
Explanation:
The fiducial points of the Celsius<span> and the </span>Fahrenheit<span> temperature </span>scales<span> are the boiling and freezing </span>points<span> of pure water at 1 atm of pressure.
In short, Your Answer would be Option D
Hope this helps!</span>
Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:
we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:
Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:
t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:
So once we got the two distances we can now find the difference between them:
so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
