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:
36 divided by (6-5) = 36
Step-by-step explanation:
6-5 = 1
36 divided by 1 = 36
D = 1/2 * (-16). That is your answer
Answer:
4x and 7x; -3 and 1: these are like terms
Step-by-step explanation:
when finding like terms, look for same variables.
9514 1404 393
Answer:
60,000
Step-by-step explanation:
To find your answer, set all other digits to zero.
60,000
is the value of the digit 6 in the 10 thousands place.
