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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.
Answer:
43.83 N
Explanation:
Given that,
The mass of an object, m = 34.4 kg
The coefficients of static and kinetic frictions for plastic on wood are 0.53 and 0.40, respectively.
The force of static friction,
The force of kinetic friction,
Net force acting on the object is :
F = 178.67-134.84
= 43.83 N
Hence, this is the required solution.