In screenshot 1 below is the hw question. That I need help with.

View an example | All parts showing Find the work required to pitch a 6.3 oz softball at 105 ft/sec. . . . The work done is equal to the kinetic energy change of the softball. Since the softball was initially at rest, its kinetic energy was 0 ft-lb. Kinetic energy is one half the product of the mass of an object times the square of the speed of the object. The relation between the weight W of an object in pounds and its mass m is W = 32 m. 2 The weight in pounds of the softball is 0.39375 lb. Since weight equals 32 times mass, the mass of the softball is 0.0123, rounded to four decimal places. The final kinetic energy of the softball is 67.8 ft-lb, rounded to the nearest tenth. Since work equals the change in kinetic energy and the initial kinetic energy was 0, the work done is 67.8 ft-lb, rounded to the nearest tenth.Find the work required to pitch a 6.5 oz softball at 100 ft/sec. The work required to pitch a 6.5 oz softball at 100 ft/sec is ft-lb. (Do not round until the final answer. Then round to the nearest tenth as needed.)