Pitching Machine Impulse In slow pitch softball, the ball (173.6 grams) is relea
ID: 1425200 • Letter: P
Question
Pitching Machine Impulse In slow pitch softball, the ball (173.6 grams) is released at approximately hip height (0.80 m) and must have an arc of at least 3 ft. The requirements for this pitching machine is to have an arc or peak ball height 1.0 m relative to the release point. The home plate is 15.2 m from the pitcher and the target height for the pitch at 15.2 m (the front of home plate) is 0.8 m.
Your assistant has performed kinematic calculations and has determined that the pitching machine needs to release the ball with a speed of 17.4 m/s toward home plate at an angle of14.7° above the horizontal plane. The ball is launched out of a tube by a pneumatic piston mechanism that accelerates a piston so that it impacts a stationary ball. Calculate the impulse (Ft) in Ns required to accelerate the softball to 17.4 m/s.(Remember that for impacts and collisions, we normally ignore the effects of gravity.)
Explanation / Answer
given
m = 173.6 g = 0.1736 kg
vi = 0
vf = 17.4 m/s
Use, Impulse-momentum theorem,
Impulse = change in momentum
= m*(vf -vi)
= 0.1736*(17.4 - 0)
= 3.02064 N.s