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:
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Step-by-step explanation:
Sector area = radius * central angle (in radians)
sector area = 22 * (5/3) PI
sector area = 36 (2/3) PI square centimeters
Answer:
not sure but can use Calculator
Right triangle
1) 75.6cm
2)10.14ft
3)16ft
Entire triangle
1)162cm
2)30.94ft
3)20ft
