As a projectile moves in its path, is there any point along the path where the velocity and acceleration vectors are:(a) perpend
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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.
(a) Because the tension in the rope would be larger than the maximum value allowed
When you are sliding down the slope, there are two forces acting on you:
- Your weight, , acting downward, where
m = 70 kg is your mass
g = 9.8 m/s^2 is the acceleration of gravity
- The tension in the rope, T, acting upward
Therefore, the equation of motion is
where a is the acceleration.
If you want to slide at constant speed, then the acceleration must be zero:
And so the equation becomes
This means that the tension in the rope must be:
Which is larger than the maximum tension allowed in the rope (500 N), so the rope will break.
(b) downward
Again, we must refer to the equation of the forces:
In this case, we want the tension in the rope to be the maximum allowed value,
T = 500 N
the other data are
m = 70 kg is your mass
g = 9.8 m/s^2 is the acceleration of gravity
Substituting into the equation, we can find the corresponding value of acceleration:
where the negative sign means the acceleration is downward.