This the final parts to my previous question ( On the surface of the moon )
The original question reads "On the surface of the moon, acceleration due to gravity is approximately 5.3 feet per second squared. Suppose a baseball is thrown upward from a height of 6 feet with an initial velocity of 15 feet per second.
A)Determine the maximum height attained by the baseball B)Determine how long it takes the ball to hit the surface of the moon. C) Find the average velocity of the ball over the time it's in flight D) Find the moment when the ball's instantaneous velocity is the same as your answer in the previous part. E) What assurance do we have that there is an answer to part D?
Part A is answered
(B) You now know that $S(t)=-2.65t^2+15t+6$, where $S(t)$ is the height above the moon’s surface at time $t$. The ball leaves your hand at time $t=0$, and it hits the moon when $S(t)=0$. Thus, all you have to do is solve $S(t)=0$ for $t$.
(C) The ball started at a height of $6$ ft and ended at a height of $0$ ft, so it travelled a net distance of how much? Be careful: this is the net distance, and the algebraic sign matters. From (B) you know how long it took. Divide the distance travelled by the time, and of course you get the average velocity $v_{\text{avg}}$.
(D) You know that the velocity at time $t$ is $v(t)=-5.3t+15$, and you know $v_{\text{avg}}$ from (C), so you just have to solve $v(t)=v_{\text{avg}}$ for $t$.
(E) HINT: Mean Value Theorem.