ground 61 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s
The speed of the ball would be 13 m/s just before it strikes the ground.
A ball is thrown horizontally from the top of a 110 m high building. 61 meters horizontally from the moment of release, the ball falls on the earth.
Velocity is defined as the displacement of the object in a given amount of time and is referred to as velocity.
First of all, we get the time of aviation from the vertical component:
Here s = 110 m and g = 9.8 m/s²
s = 1/2 × gt²
Solve for t
t = √(2s/g)
Substitute the values of s = 110 m and g = 9.8 m/s² in the equation,
t = √(2×110/9.8)
t = √22.44
t = 4.7 sec
The horizontal component of velocity is constant :
v = s/t
Here s = 61 m and t = 4.7 sec
v = 61/4.7
v = 12.97 ≈ 13
Therefore, the speed of the ball would be 13 m/s just before it strikes the ground.
Learn more about Velocity here:
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Ive been struggling
Answer:
The answer to your question is Area = 9 3/8
Step-by-step explanation:
Data
length = 3 3/4 cm
width = 2 1/2 cm
Process
1.- Write the equation of the Area of a rectangle
Area = length x width
2.- Convert the values to improper fractions
3 3/4 = (12 + 3)/ 4 = 15/4
2 1/2 = (4 + 1) / 2 = 5/2
3.- Substitution
Area = (15/4)(5/2)
4.- Simplification
Area = 75/8
5.- Convert to a mixed fraction
9
8 75
3
6.- Result
Area = 9 3/8
B.) $9,841.16
C.) $7,353.88
Answer:
its b
im on plato and got it right
Screen Shot 2020-05-18 at 9.57.43 AM
Answer:
there is nothing there
Step-by-step explanation:
literally
Answe
Step-by-step explanation:
Answer:
-1.2
Step-by-step explanation:
solve