A ball is shot from the ground straight up into the air with initial velocity of 47 ft/sec. Assuming that the air resistance can be ignored, how high does it go? Note: The velocity of the ball is 0 at the top of it's flight. Find v(t), then set v(t)=0. Then find the function for the position of the ball, s(t). Use this and the value found from v(t)=0 to complete the problem. Hint: The acceleration due to gravity is a(t)=-32 ft per second squared.