Complete question:
A student throws a heavy ball downward off the top of a building with a speed of 18m/s. The ball reaches a speed of 41m/s just before striking the ground. Neglect drag, find the height of the building.
Answer:
The height of the building is 69.235 m
Step-by-step explanation:
Given;
initial velocity of the ball, u = 18 m/s
final velocity of the ball, v = 41 m/s
The height of the building is equal to distance traveled by the ball downward.
Apply the following kinematic equation;
v² = u² + 2gh
where;
g is acceleration due to gravity
h is height of the building
41² = 18² + 2(9.8)h
1681 = 324 + 19.6h
19.6h = 1681 - 324
19.6h = 1357
h = 1357 / 19.6
h = 69.235 m
Therefore, the height of the building is 69.235 m
Answer:
n= -0.9
Step-by-step explanation:
-4.2n-1.7=2.08 ( move the constant to the right)
-4.2n = 2.08 + 1.7( calculate)
n = -0.9
Answer:
D. 75(5r+3d+4f) and 75(5r)+75(3d)+75(4f)
In (5,0), five represents how many times he will move over in the X axis. Since it’s zero, you do not move at all. However you will move five spot up, because the five is on the Y axis. Then, he will do the same thing with your second pair of coordinates. You do not move on the X axis, but you will move two spots down on the Y axis. Then, trace your pencil between the two dots. You will notice that it passes seven intersections, so your answer will be seven units.
4 is the proportion because 8 + 4 is 12 then keep adding 12 + 4 = 16 + 4 = 20 So that mean it is proportional I hope that helps.
