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avanturin [10]
3 years ago
14

A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in t

he pitcher's hand is 35.0 m/s and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm?
Group of answer choices
A) 10.5 rad/s
B) 0.0952 rad/s
C) 0.00857 rad/s
D) 117 rad/s
Physics
1 answer:
stiv31 [10]3 years ago
3 0

Answer:

D) 117 rad/s

Explanation:

We can treat this system as a circular motion where the origin is the elbow joint, the ball rotation velocity v is 35 m/s, the rotation radius is r = 0.3m.

As the ball is leaving the pitcher hand at such speed and rotation radius. Its angular velocity is:

\omega = \frac{v}{r} = \frac{35}{0.3} = 117 rad/s

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Answer:

The pitching speed of your friend is 33.20 m/s

Explanation:

<em>Lets explain how to solve the problem</em>

Your friend throw the ball horizontally that means the vertical initial

component of velocity is zero (u_{y}=0).

The ball is thrown from a height 4 meters above the ground.

The height h=u_{y}t+\frac{1}{2}gt^{2}

<u><em>Remember:</em></u> the height is negative value because its below the point of

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h = -4 m , u_{y}=0 and g = -9.8 m/s²(downward)

<em>Substitute these values in the rule above</em>

⇒ 4=0-\frac{1}{2}(9.8)t^{2}

⇒ -4 = -4.9t² (multiply both sides by -1)

⇒ 4 = 4.9t² (divide both sides by 4.9)

⇒ 0.81633 = t² (take √ for both sides)

⇒ <em>t = 0.9035</em>

Then the time of the ball to land on the ground is 0.9035 seconds

The range of the ball on the ground is 30 m

The range R=u_{x}*t, where u_{x} is the horizontal

component of the initial velocity

R = 30 meters and t = 0.9035

⇒ 30=u_{x}(0.9035) (divide both sides by 0.9035)

⇒ u_{x}=33.20 m/s

<em>The pitching speed of your friend is 33.20 m/s </em>

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