For the second question,
The ball is moving downward from a certain height to ground zero.
So, the initial velocity of 40 feet/sec will be the final velocity in the downward direction.
Because, when you throw a ball upward vertically with an initial velocity of u, at the highest point, the initial velocity becomes zero (or, momentarily rest) and then, it travels downward with final velocity v, but having the same magnitude as u.
So, using the following equations, one can find the solution to the question (b).
Step 1) Find the time taken to reach the maximum height that the ball will reach vertically upward.
Step 2) Once it has reached that height, then find the time taken to reach the ground.
So, we know, v = u + at. In this context, a = g (gravitational acceleration).
So, for Step 1) t1 = u sin(theta) / g = u/g [ v = u Sin(theta) - gt1 ; since v = 0, therefore, t1 = u sin(theta) / g ] [Here, theta = 90, since the ball is thrown vertically upward at an angle of 90 degree with the horizontal].
And for Step 2) t2 = u sin(theta) / g = u/g [ v = u Sin(theta) + gt2 ; since v = u sin(theta) and u = 0, therefore, t2 = u sin(theta) / g ] [Here, theta = 90, since the ball is thrown vertically upward at an angle of 90 degree with the horizontal].
Therefore, the Total time of flight for the ball after being hit is as follows.
t = t1 + t2 = (2*u)/g [ g = 32.1741 feet per second squared]
= (2*40)/(32.1741)
= 2.4865 seconds.