Final answer:
The force on the 0.150 kg baseball at both half its maximum height and its peak during an upward throw is constant and equals its weight. The calculation, using F=mg for the force of gravity, results in a force of 1.47 N.
Step-by-step explanation:
The force on a 0.150 kg baseball when thrown upwards at an initial speed of 20.0 m/s is solely due to gravity at any point during its ascent or descent, until it reaches the ground or is caught. This force, often called weight, is constant and equals the mass of the ball multiplied by the acceleration due to gravity (F = mg), regardless of the ball's position in its trajectory. When the ball is at half its maximum height or at its peak, the only force acting on it is still its weight, and it is not dependent on the ball's velocity or height. Therefore, the force exerted by gravity on the baseball is the same at both half its maximum height and its peak.
Calculated using the formula F = m \( \times \) g, where m is the mass of the baseball (0.150 kg) and g is the acceleration due to gravity (approximately 9.8 m/s2), the force of gravity on the baseball is:
F = 0.150 kg \( \times \) 9.8 m/s2 = 1.47 N