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
47 times 22 = 1034
1034/2 = 517
380 is less than 517 so Renee has more than enough for each student.
Answer:
that's a
Step-by-step explanation:
y=[x]-3
I think
Lets divide the problem in parts, first for sandwiches she used 3/4 out of 20 ounces of ham:
(3/4)(20) = 60/4 = 15
so for sandwiches Lauren used 15 ounces
Then for omelete she used 1/5, that is:
(1/5)(20) = 20/5 = 4
then Lauren used 4 ounces for omelete
So Lauren used 15 ounces for sandwiches and 4 ounces for omelete, that is 19 ounces in total, so there are left from the 20 original ounces:
20 - 19 = 1
there is 1 ounce of ham left
