Final answer:
To find the maximum height the ball reaches, we set the velocity function derived from s(t) to zero and solve for time. Then, we substitute that time into the s(t) function to find the maximum height, which yields a height of 9.25 feet, although this does not match any of the multiple-choice options provided.
Step-by-step explanation:
The problem presented asks for the maximum height a ball will reach when thrown directly upwards from a starting height of 3 feet with an initial velocity of 20 ft/sec. Using the position function s(t) = -16t² + 20t + 3, we need to determine the time at which the ball reaches its peak height. At peak height, the velocity of the ball will be zero.
Using the velocity function which is the derivative of the position function, v(t) = s'(t) = -32t + 20, setting v(t) to zero gives us the time when the ball's velocity is zero and thus at its highest point: 0 = -32t + 20. Solving this, we get t = 20/32 or t = 5/8 seconds.
Now, substituting this time back into the position function, s(5/8) = -16(5/8)² + 20(5/8) + 3 yield the maximum height: s(5/8) = -16(25/64) + 20(5/8) + 3, which simplifies to s(5/8) = -6.25 + 12.5 + 3 and further to s(5/8) = 9.25 ft. However, the question does not include this value, potentially indicating a typo or mistake in the question's multiple-choice options. The closest answer is 9 ft, but as it stands, none of the provided options A) 20ft B) 23ft C) 25ft D) 28ft are correct.