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
You multiply 33 by 0.9 because it's minus 10% (0.1) and the answer is $29.7
Answer:
Step-by-step explanation:
84% Vol =
3
x base area x ht
The pyramid below has a rectangular base and E
is directly above A.
Answer:
i think its .625
Step-by-step explanation:
all you need to do to make a fraction into a decimal is divide the nominator by the denominator
Answer:
Step-by-step explanation:
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