Question 1. Recall the ball-rolling-down-a-hill example we saw in class: (orb) b ( as ) If a stopped ball at the top (namely F(0) = (0, b) ) rolls down the hill, after t seconds, the ball will be at 4.9abt- 4.9632 F(t) = (x(t), y(0)) = a2 + 62 (1) Show that' $10(1)12 + 9.8y(1) = =10(0)12 +9.8y(0). Here, O(t) is the velocity function, given by 7(t) = F'(t), and y(t) is the y-component of the position function f(t) (namely, y(t) = b - 4. (2) Compute the time that the ball A takes to roll down the hill, when a = b = 20.(3) Suppose the ball B starts at the same position, (0, 20), but takes a different path: This relation is an instance of conservation of energy. and is a very fundamental rule in physics. Namely, this ball first rolls down a steeper hill, and then rolls on the ground. Suppose that, at the end of the hill, the ball keeps its speed while changing its direction. Compute the time that the ball B takes to reach the point (20, (). (4) Show that, if the ball A and B start rolling at the same time, the ball B reaches the goal point (20, 0) faster than the ball A. Show also that the ball B travels more distance than the ball A. This means that the shortest path might not be the fastest path