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Sagot :
Free-fall
Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration .
Free-fall motion is said to have a constant acceleration so that the change in velocity at a certain time interval is constant. By using the free-fall motion formula, the position and the velocity for the coin :
A. 1.0 s : h = 4.9 m , v = 9.8 m/s
B. 2.0 s : h = 19.6 m, v = 19.6 m/s
C. 3.0 s : h = 29.4 m, v = 44.1 m/s
Further explanation
The motion of a ball dropped at a point is an example of free-fall motion
Free-fall motion is included in uniformly accelerated motion with constant acceleration. Its acceleration is influenced by Earth's gravity (g=10 m/ s²)
In this motion, the equation applies
- h = vot + 1/2 gt²
- vt = vo + gt
The initial speed of the ball when released is 0, so :
- vt = gt
- h = 1/2 gt²
h = distance / height
g = acceleration due to gravity
t = time
vt = speed when t seconds
Given
g = 9.8 m/s²
time : A) 1.0 s B) 2.0 s C) 3.0 s
Required
position and velocity
Solution
- A. 1.0 s
vt = gt = 9.8 x 1 = 9.8 m/s
h = 1/2 gt² = 1/2 x 9.8 x 1² = 4.9 m
- B. 2.0 s
vt = gt = 9.8 x 2 = 19.6 m/s
h = 1/2 gt² = 1/2 x 9.8 x 2² = 19.6 m
- C. 3.0 s
vt = gt = 9.8 x 3 = 29.4 m/s
h = 1/2 gt² = 1/2 x 9.8 x 3² = 44.1 m
Learn more
The velocity of the ball just before it hits the ground : https://brainly.ph/question/519262
The speed or velocity of the ball when it reaches its maximum height : https://brainly.ph/question/106747
Graphs of distance vs time : https://brainly.ph/question/512733
#LearnWithBrainly
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