Final answer:
Using conservation of energy principles and given an initial velocity of 20 m/s upwards for a ball thrown straight up, we find the maximum height reached by the ball to be approximately 20.4 meters. The provided options do not match this result, indicating a potential error in the test question.
Step-by-step explanation:
If the quarterback accidentally threw the ball straight upwards and all of the kinetic energy turned into potential energy at the top of its path, we can use the conservation of energy principle to determine how high the ball would go. The formula for gravitational potential energy at a height h is PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s²), and h is the height. We equate this to the kinetic energy, which is given by KE = 1/2 mv², where v is the velocity of the ball.
In this specific scenario, if a ball is thrown straight up with an initial velocity of 20 m/s upwards, to find the maximum height h, we would set the kinetic energy equal to the potential energy at the top of the ball's path:
1/2 mv² = mgh.
Given that m would cancel out on both sides of the equation, we solve for h:
h = (1/2 v²) / g
h = (1/2 × 20²) / 9.8
h = (1/2 × 400) / 9.8
h = 200 / 9.8
h ≈ 20.4 m
Therefore, the correct answer is that the ball would reach a max height of approximately 20.4 meters. However, none of the given options matches this result. If this occurred on an actual test question, it would likely be an error in the question design, and it would be best to clarify with the instructor.