fastest measured pitched baseball left the pitchers hand at speed of 45 m/s. if the pitcher was in contact with the ball at a disturbance of 1.50 m and produced constant acceleration.
a.what accelerate did he gave the ball?
b.how much time did it take him to pitch?
To find the answers to the given questions, we can use the equations of motion. One of the most relevant equations is the equation for linear motion in terms of acceleration, final velocity, initial velocity, and displacement:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
Let's solve each part of the question:
a. To find the acceleration, we need the final velocity (v), initial velocity (u), and displacement (s). The initial velocity is the speed at which the ball left the pitcher's hand, so u = 0 m/s. The final velocity is given as v = 45 m/s. The displacement is the distance the pitcher moved while in contact with the ball, which is given as s = 1.50 m.
Using the equation v^2 = u^2 + 2as, we can rearrange it to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
Substituting the values we have, we get:
a = (45^2 - 0^2) / (2 * 1.50)
a = 2025 / 3
a ≈ 675 m/s^2
Therefore, the pitcher gave the ball an acceleration of approximately 675 m/s^2.
b. To find the time it took for the pitcher to pitch, we can use another equation of motion that relates acceleration, time, initial velocity, and displacement:
s = ut + (1/2)at^2
Since the initial velocity is 0 m/s and the displacement is given as 1.50 m, we can rearrange the equation to solve for time (t):
t = sqrt((2s) / a)
Substituting the values we have, we get:
t = sqrt((2 * 1.50) / 675)
t = sqrt(3/675)
t ≈ sqrt(0.0044)
t ≈ 0.066 s
Therefore, it took the pitcher approximately 0.066 seconds to pitch.
Note: These calculations assume constant acceleration and may not accurately represent the actual physics of pitching a baseball.
Vf^2=vi^2+2ad solve for a.
b.
force*time=massball*45
massball*a*time=massball*45
solve for time.