Final answer:
The correct quadratic function representing the height above ground of the ball thrown upward is h(t) = -16t^2 + 64t + 48. The equation takes into account the starting height of the ball, the effect of gravity over time, and conforms to the given points that the ball is at 64 feet at 1 second and hits the ground after 3 seconds.
Step-by-step explanation:
The student is asked to write the equation of the quadratic function representing the height above ground of a ball thrown upward from a building. We are given that the ball is at 64 feet at t = 1 second and it hits the ground after 3 seconds. Since we know the ball starts at 48 feet and reaches the ground in 3 seconds, the equation will have a root at t = 3, meaning the term (t - 3) will be a factor of the quadratic equation.
Using the information that the ball is at 64 feet after 1 second, we can calculate that the coefficient for the t2 term must be -16, because in one second, the change from the initial height of 48 feet to 64 feet is an increase of 16 feet. This change can be accounted for by an initial positive velocity component that is being affected by the acceleration due to gravity, which is approximately -32 feet/second2 (or -16 feet/second2 if using the physics convention where "g" is taken as positive for free-fall problems). Since the ball's velocity decreases by 32 feet/second each second (64 feet per 2 seconds), after the first second, the ball would rise 16 feet above its initial height, giving us 64 feet.
Therefore, after considering the factors and the information given, the correct quadratic equation will be of the form h(t) = -16t2 + vt + 48, where v is the initial velocity. The only equation that fits all the given information is h(t) = -16t2 + 64t + 48.