The forces exerted by Earth and a skier become an action-reaction force pair when the skier pushes the ski poles against Earth.
1 answer:
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Answer:
V1 =8.1 m/s
Explanation:
height at highest point (h2) = 4.1 m
height at lowest point (h1) = 0.8 m
acceleration due to gravity (g) = 9.8 m/s^{2}
from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore
mgh1 + = mgh2 +
where
- speed at highest point = V2
- speed at lowest point = V1
- mass of the girl and swing = m
- at the highest point, the speed is minimum (V1 = 0)
- at the lowest point the speed is maximum (V2 is the maximum speed)
- therefore the equation becomes mgh1 +
= mgh2
m(gh1 + ) = m(gh2)
gh1 + = gh2
V1 =
now we can substitute all required values into the equation above.
V1 =
V1 =
V1 =8.1 m/s
Answer:
The initial velocity of the pitch is approximately 36.5 m/s
Explanation:
The given parameters of the thrown fastball are;
The height at which the pitcher throws the fastball, h₁ = 2.65 m
The angle direction in which the ball is thrown, θ = 2.5° below the horizontal
The height above the ground the catcher catches the ball, h₂ = 1.02 m
The distance between the pitcher's mound and the home plate = 18.5 m
Let 'u' represent the initial velocity of the pitch
From h = ·t + 1/2·g·t², we have;
= The vertical velocity = u·sin(θ) = u·sin(2.5°)
h = 2.65 m - 1.02 m = 1.63 m
uₓ·t = u·cos(θ) = u·cos(2.5°) × t = 18.5 m
∴ t = 18.5 m/(u·cos(2.5°))
∴ h = ·t + 1/2·g·t² = (u·sin(2.5°))×(18.5/(u·cos(2.5°))) + 1/2·g·t²
1.63 = 8.5·tan(2.5°) + 1/2 × 9.8 × t²
t² = (1.63 - 8.5·tan(2.5°))/(1/2 × 9.8) = 0.25691469087
t = √(0.25691469087) ≈ 0.50686752763
t ≈ 0.50686752763 seconds
u = 18.5 m/(t·cos(2.5°)) = 18.5 m/(0.50686752763 s × cos(2.5°)) = 36.5334603 m/s ≈ 36.5 m/s
The initial velocity of the pitch = u ≈ 36.5 m/s.