Answer:
you have to be more specific on what your talking about
Step-by-step explanation:
D.
The equation would look like this
62,000(1+.04)^6
Answer:
D
Step-by-step explanation:
Sum up, subtract and simplify
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:
61
Step-by-step explanation:
The complete angle is 90, a part of it is 29, so the rest is 90 - 29
