Final answer:
The time at which the ball stops going up and starts to descend is when its velocity equals zero. By setting the derivative of the height function equal to zero, we find that this occurs at approximately 19.90625 seconds. However, this does not match any of the provided options, indicating a possible typo in the question.
Step-by-step explanation:
To determine when the ball stops going up and starts returning to the ground, we need to find the time when its vertical velocity, which is the derivative of the height function h(t), is zero. The height function given is h(t) = 637t - 16t2. The first derivative of the height function with respect to time t, which represents velocity, is v(t) = h'(t) = 637 - 32t. Setting this equal to zero will give us the time when the vertical velocity is zero and the ball starts to fall back to the earth.
Solving for t from v(t) = 0, we get:
0 = 637 - 32t
t = 637 / 32
t = 19.90625 seconds
However, this value does not match any of the options provided in the question. The ball's stopping point in the air corresponds to the peak of the parabola represented by the height function, and we typically calculate it by finding the vertex of the parabola. In fact, the coefficient of the t2 term should be negative; hence there might be a typo in the function provided. If corrected to h(t) = 637t - 16t2, then the vertex can be calculated as t = -b/(2a), where a is the coefficient of t2 and b is the coefficient of t. Substituting the values, we get t = (-637)/(2*(-16)), which simplifies to t = 637/32, giving us the same result as before.
Therefore, there might be a mistake in the given height function, or the options provided do not match the correct calculations. It's important to verify the accuracy of the function and the options for time given in the question.