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
Here is the solution based on the given situation above.
Given: 450 kg = weight limit of the elevator
750 kg = current weight of the group of passengers
? = number of passengers who weigh 70 kgs that need to get off the elevator
So to get the answer, first we need to deduct 450kg from 750 kgs to get the excess weight. So 750 - 450 is 300kgs. Next, we divide 300 by 70kgs to get the number of passengers that need to get off so 300/70 is 4.3. Therefore, there should be 4 passengers who weigh 70 kilograms that need to get off the elevator.
Answer:
I feel like its 75 degrees and also are u in 8th and have u done power quotient and product rule before if so I need help
The answer is (a) 2x^4 y^8
