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.