7. How long does it take a ball rolling down a hill to change its speed from 3 m/sec to 34.5 m/sec
1 answer:
The time elapsed is 9 seconds
Explanation:
The motion of the ball is a uniformly accelerated motion (a motion with constant acceleration), so we can use the following suvat equation:
where :
v is the final velocity of the ball
u is the initial velocity
a is the acceleration
t is the time elapsed
For the ball in this problem, we have:
u = 3 m/s is the initial velocity
v = 34.5 m/s is the final velocity
is the acceleration
Solving for t, we find the time taken for this change in velocity:
Learn more about acceleration:
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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.