calculate the horizontal force needed to make a 3.8 kg baseball accelerate at a 641m/s^2

Question

is part of a worksheet and I don't know how to solve it and need help

answer the problem by using the equation for force (F=ma) and (W=mg)

Answer 1

Consider an analysis of this problem with realistic quantities.

[The fastest recorded baseball pitch is 105.8 miles per hour or 47.30629756 meters per second.

Regulation distance from the pitcher's mound to home plate is 60.5 feet or 18.4440905 meters.

A regulation baseball has a mass of some 5 ounces or 0.145 kilograms. (The figure of 3.8 kilograms 

given would translate to a baseball weight of 8.398 pounds.)

The given acceleration of 641 meters-per-second-squared would mean an increase in speed of 

1433.589257 miles per hour for every second in flight.]

Here a pitcher can accelerate with constant horizontal force a baseball with a mass of 0.145 kg

from rest to a speed of 47.30629756 meters per second in a distance of 18.4440905 meters.

Evaluate constant acceleration from v² − v₀² = 2ax; place values to obtain

 47.30629756² (m/s)² − 0² (m/s)² = 2a (m/s²) × 18.4440905 (m).  

Simplify this last to 47.30629756² (m/s)² = 2a(18.4440905) (m/s)² or acceleration is equal to 

(2237.885789 divided by 2 divided by 18.4440905) or 60.6667428 m/s².          

Then take the horizontal force sought from Force Equals Mass Times Acceleration or

F = (0.145 kg) • (60.6667428 m/s²) which goes to 8.796677705 Newtons.

8.796677705 Newtons (or 1.9775718244343 earth-pounds of force) is 6.190483958 times the gravitational force on the baseball (equal to 0.145 kilograms times 9.8 meters-per-second-squared or 1.421 Newtons) so the effect of gravity on the baseball is greatly diminished at such an acceleration.