There is one degree of freedom
Assume that the initial velocity of the throw is $p{\bf i} + q{\bf j}$, so $u_x=p$ and $u_y=q$. The acceleration due to gravity is $0{\bf i} - g{\bf j}$ , meaning that $a_x=0$ and $a_y=-g$. The throw point is 15ft from the basket, so $s_x=15$. The initial throw height is 8ft above the ground and the basket height is 10ft, so $s_y=2$.
This takes the throw point as the origin.
Considering only vertical motion: $s=ut+\frac{1}{2}at^2$ gives $15 = qt-\frac{1}{2}gt^2$.
Considering only horizontal motion: $s=ut+\frac{1}{2}at^2$ gives $2 = pt$.
Solving $2=pt$ gives $t=\frac{2}{p}$, assuming that $p \neq 0$, and so $15=q\left(\frac{2}{p}\right)-\frac{1}{2}g\left(\frac{2}{p}\right)^2$
$$15p^2=2pq-2g$$
$$q=\frac{15p^2+2g}{2p}$$
For any choice of $p$, you get a $q$ which gives the initial velocity.
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.
Best Answer
You've ignored the multiplication of initial velocity by time of flight. The total distance above ground is given by $$3+14t-5t^2$$which then must be solved for equality to zero distance above ground, and account for common sense.
Gravitational acceleration creates a change in distance of $$-5t^2$$ if we use 1 significant figure. The only thing affecting the gravitational changes is gravity itself, which has no dependence on velocity (at least in his scale).
The initial velocity changes distance by a separate factor,$$v_it$$ which does not depend on gravity, hence these terms are separate in the equation. The constant term is straightforwardly the height at time zero.