Solve 1-D Kinematics Problem: Baseball Pitcher's Acceleration

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A baseball pitcher throws a fastball at 36.1 m/s over a distance of 3.06 m, requiring the calculation of uniform acceleration. The initial velocity is 0 m/s since the ball is held before being thrown. Using the kinematic equation, the time to reach the final velocity is determined to be approximately 0.329 seconds. Substituting this back into the acceleration formula yields an acceleration of 56.9 m/s². Proper application of kinematic equations is essential for solving such physics problems effectively.
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I need some help with this problem for phys class

A baseball pitcher throws a fastball at the speed of 36.1 m/s. the acceleration occurs as the pitcher holds the ball in his hand and moves it through an almost straight-line distance of 3.06m. calculate the acceleration, assumming its uniform.

i have exhausted my resources trying to figure out how to go about solving this problem, help would be appreciated
 
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mgiddy911 said:
I need some help with this problem for phys class

A baseball pitcher throws a fastball at the speed of 36.1 m/s. the acceleration occurs as the pitcher holds the ball in his hand and moves it through an almost straight-line distance of 3.06m. calculate the acceleration, assumming its uniform.

i have exhausted my resources trying to figure out how to go about solving this problem, help would be appreciated

use vf^2 = vi^2 + 2ad
 
.

To solve this problem, we can use the formula for acceleration, which is a = (v2 - v1)/t, where v2 is the final velocity, v1 is the initial velocity, and t is the time. In this case, we are given the final velocity (36.1 m/s) and the distance (3.06 m) but we need to find the initial velocity and time.

To find the initial velocity, we can use the fact that the pitcher holds the ball in his hand before throwing it, so the initial velocity is 0 m/s.

Now, we can rearrange the formula to solve for time: t = (v2 - v1)/a. Plugging in the values we know, we get t = (36.1 m/s - 0 m/s)/a.

Since we are assuming the acceleration is uniform, we can use the distance formula, d = v1t + 1/2at^2, to solve for the acceleration. Plugging in the values we know, we get 3.06 m = 0 m/s * t + 1/2 * a * t^2.

Simplifying, we get 3.06 = 1/2 * a * t^2. We can rearrange this to solve for a: a = 2 * (3.06/t^2).

Now, we can substitute this value for a back into our equation for time: t = (36.1 m/s - 0 m/s)/(2 * (3.06/t^2)).

Simplifying, we get t = 0.329 seconds.

Therefore, the acceleration is a = 2 * (3.06/0.329^2) = 56.9 m/s^2.

So, the pitcher's acceleration is 56.9 m/s^2. I hope this helps to solve the problem for your physics class. Remember to always use the appropriate formulas and units when solving kinematics problems.
 
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