Hi: I need help with the following physics problem. Can you explain it to me? Thank you,
The fastest measured pitched baseball left the pitcher's hand at a speed of 50.0 . If the pitcher was in contact with the ball over a distance of 1.50 and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?
Of course! Let's break down the problem step by step.
We are given the initial velocity of the ball, v0 = 0 (since it starts from rest), the final velocity, v = 50.0 m/s, and the distance traveled by the pitcher's hand, d = 1.50 m.
(a) To find the acceleration, we can use the kinematic equation that relates velocity, distance, and acceleration:
v^2 = v0^2 + 2aΔd
Plugging in the values we know, we have:
(50.0 m/s)^2 = 0^2 + 2a(1.50 m)
2500 m^2/s^2 = 3a
Now, we can solve for a:
a = (2500 m^2/s^2) / 3 ≈ 833.33 m^2/s^2
So, the acceleration given by the pitcher is approximately 833.33 m^2/s^2.
(b) To find the time it took for the ball to travel the given distance, we can use another kinematic equation:
d = v0t + (1/2)at^2
Since the initial velocity is 0 m/s, the equation simplifies to:
d = (1/2)at^2
Plugging in the values we know, we have:
1.50 m = (1/2)(833.33 m^2/s^2) t^2
Rearranging and solving for t, we get:
t^2 = (2 * 1.50 m) / 833.33 m^2/s^2
t^2 ≈ 0.0036 s^2
t ≈ √(0.0036 s^2) ≈ 0.06 s
Therefore, it took approximately 0.06 seconds for the pitcher to pitch the ball.
To summarize:
(a) The acceleration given by the pitcher is approximately 833.33 m^2/s^2.
(b) It took approximately 0.06 seconds for the pitcher to pitch the ball.
I hope this helps! Let me know if you have any further questions.