Thank you for helping. We are not using double integrals yet. I am having a problem setting up the equation. I know that the integral of acceleration gives velocity, and velocity integral gives position. I am not sure how or what to do with the values given. Thank you again.

(1) A ball is thrown upward with a speed of 48 ft/s from the edge of a cliff 432 ft above the ground. The motion is vertical and we choose the positive direction to be upward. At time t the height in feet above the ground is s(t) and the velocity v(t) in ft/s is decreasing. Therefore, the acceleration a(t) in ft/s2 must be negative. Given that a(t) = -32. (i) Find its height above the ground t seconds later. Hints: Given that v(0) = 48. To find v(t), use v(t) = v(0) + So a(x) dx. Given that s(0) = 432. To find s(t), use s(t) = s(0) + fv(x)dx. (ii) When does it reach its maximum height? Hint: The maximum height is reached when v(t) = 0. (iii) When does it hit the ground? Hint: The ball hits the ground when s(t) = 0