In slow pitch softball, the ball (173.6 grams) is released at approximately hip

Question

In slow pitch softball, the ball (173.6 grams) is released at approximately hip height (0.80 m) and must have an are 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 theta of 14.7 degree 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 (F Delta t) in N s required to accelerate the softball to 17.4 m/s. (Remember that for impacts and collisions, ne normally ignore the effects of gravity.)

Explanation / Answer

I don't understand why so many data.

p=mv=0.1736*17.4=3.02064 Ns.

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