The acceleration of a baseball pitcher's hand as he delivers a pitch is extreme. For a professional player, this acceleration phase lasts only 50 ms, during which the ball's speed increases from 0 to about 90 mph, or 40 m/s.

What is the force of the pitcher's hand on the 0.145 kg ball during this acceleration phase?

Answers

Answer 1
Final answer:

The force by the pitcher's hand on the ball during the acceleration phase is found using Newton's Second Law of Motion and is calculated to be 116 Newton.

Explanation:

The force exerted by the pitcher's hand can be found using Newtons Second Law of Motion which states that the force acting on an object is equal to the mass of the object times its acceleration.

it can be expressed as

F = m * a

Here, the mass (m) is 0.145 kg, and the acceleration (a) can be found using the formula a = Δv/Δt, where Δv is the change in velocity (40 m/s) and Δt is the change in time (50 ms or 0.05 s).

So, a = 40/0.05 = 800 m/s², and then the force F = 0.145 * 800 = 116 N. Therefore, the force of the pitcher's hand on the ball during this acceleration phase is 116 Newton.

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Answer 2
Final answer:

The force of the pitcher's hand on the ball can be calculated using Newton's second law of motion.

Explanation:

The force of the pitcher's hand on the ball can be calculated using Newton's second law of motion, which states that force equals mass times acceleration. In this case, the mass of the ball is 0.145 kg and the acceleration is the change in velocity divided by the time interval. The change in velocity is 40 m/s (the final velocity) minus 0 m/s (the initial velocity), and the time interval is 50 ms (or 0.05 s). Therefore, the force can be calculated as:

Force = mass × acceleration = 0.145 kg × (40 m/s - 0 m/s) / 0.05 s = 0.116 N.

So, the force of the pitcher's hand on the ball during this acceleration phase is approximately 0.116 Newtons.

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Related Questions

A satellite with a mass of 5.6 E 5 kg is orbiting the Earth in a circular path. Determine the satellite's velocity if it is orbiting at a distance of 6.8 E 5 m above the Earth's surface. Earth's mass = 5.98 E 24 kg; Earth's radius = 6.357 E 6 m.

A) 6,800 m/s
B) 7,200 m/s
C) 7,500 m/s
D) 7,900 m/s

Answers

Answer:

C) 7,500 m/s

Explanation:

The satellite's acceleration due to gravity equals its centripetal acceleration.

v² / r = GM / r²

Solving for velocity:

v² = GM / r

v = √(GM / r)

Given:

G = 6.67×10⁻¹¹ m³/kg/s²

M = 5.98×10²⁴ kg

r = 6.357×10⁶ m + 6.8×10⁵ m = 7.037×10⁶ m

Substituting the values:

v = √(6.67×10⁻¹¹ × 5.98×10²⁴ / 7.037×10⁶)

v = √(5.67×10⁷)

v = 7500 m/s

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