A fast, measured pitched baseball left the pitcher's hand at a speed of 48.6 m/s. The pitcher was in contact with the ball over a distance of 1.44 m and produced constant acceleration.
I know that t=(Vf-Vi)/a based on a kinematic equation. Also, I setup:
1.44=Vi(t) + 0.5at^2
I'm mainly confused on what the initial and final velocities would be. Would the initial be 0 or 48.6 m/s?
In this scenario, the initial velocity (Vi) refers to the velocity when the ball is released from the pitcher's hand, before any acceleration occurs. The final velocity (Vf) refers to the velocity of the ball when it reaches the end of the pitcher's arm, after the acceleration has taken place.
Given that the ball is initially at rest in the pitcher's hand and then accelerates to a certain velocity, the initial velocity (Vi) would be 0 m/s. The final velocity (Vf) would be the velocity at the end of the arm, which is 48.6 m/s.
So, when applying the kinematic equation 1.44 = Vi(t) + 0.5at^2, the equation becomes:
1.44 = 0(t) + 0.5at^2
Since the initial velocity (Vi) is 0, the first term in the equation becomes 0, and you are left with:
1.44 = 0.5at^2
Now, you have the equation without the initial velocity.
To determine the initial and final velocities in this scenario, we can use the information given.
The initial velocity refers to the velocity of the baseball as it leaves the pitcher's hand. In this case, the given information states that the baseball left the pitcher's hand at a speed of 48.6 m/s. Therefore, the initial velocity (Vi) is indeed 48.6 m/s.
The final velocity (Vf) is the velocity of the baseball at the end of the given distance. In this case, since the baseball is pitched and leaves the pitcher's hand, we can consider the final velocity to be 0 m/s. This is because the pitcher's contact with the ball ends at this point, and assuming no external forces act on the ball during the given distance, it will eventually slow down and come to a stop.
Therefore, in the kinematic equation that you set up, with Vi = 48.6 m/s and Vf = 0 m/s, you can use the equation:
1.44 = Vi(t) + 0.5at^2
Substituting the values:
1.44 = (48.6)(t) + 0.5a(t^2)
You can now solve this equation for the time (t) and the acceleration (a) using known values and solving for the unknowns.