Question

1. when throwing a softball directly upward from a height of 5 ft with an initial velocity of 50 ft/sec, the height of the softball after t seconds is given by (y(t)=-16t^{2}+50t + 5) (until the ball hits the ground). a. find the velocity (y(t)). b. over what time interval is the ball going upward? c. at what time does the ball reach its maximum height? d. what is the maximum height reached by the ball? 7. determine all the points on the graph of (y = x^{3}+x^{2}-x + 1) where the tangent is horizontal.

Explanation:

Step1: Differentiate \(y(t)\) to find \(y'(t)\)

Using the power - rule \((x^n)'=nx^{n - 1}\), if \(y(t)=-16t^{2}+50t + 5\), then \(y'(t)=-32t + 50\).

Step2: Find when the ball is going upward

The ball is going upward when \(y'(t)>0\). Solve the inequality \(-32t + 50>0\).
\(-32t>-50\), so \(t<\frac{50}{32}=\frac{25}{16}\). Since \(t\geq0\), the time interval is \(0\leq t<\frac{25}{16}\) seconds.

Step3: Find the time of maximum height

The ball reaches its maximum height when \(y'(t) = 0\). Set \(-32t+50 = 0\).
\(-32t=-50\), then \(t=\frac{50}{32}=\frac{25}{16}=1.5625\) seconds.

Step4: Find the maximum height

Substitute \(t = \frac{25}{16}\) into \(y(t)\).
\[

$$\begin{align*} y(\frac{25}{16})&=-16(\frac{25}{16})^{2}+50(\frac{25}{16})+5\\ &=-16\times\frac{625}{256}+ \frac{1250}{16}+5\\ &=-\frac{625}{16}+\frac{1250}{16}+5\\ &=\frac{-625 + 1250}{16}+5\\ &=\frac{625}{16}+5\\ &=\frac{625+80}{16}\\ &=\frac{705}{16}=44.0625 \end{align*}$$

\]
For question 7:

Step1: Differentiate \(y = x^{3}+x^{2}-x + 1\)

Using the power - rule, \(y'=3x^{2}+2x - 1\).

Step2: Set \(y' = 0\) for horizontal tangents

Factor \(3x^{2}+2x - 1=(3x - 1)(x + 1)=0\).

Step3: Solve for \(x\)

If \(3x-1 = 0\), then \(x=\frac{1}{3}\); if \(x + 1=0\), then \(x=-1\).
When \(x=\frac{1}{3}\), \(y=(\frac{1}{3})^{3}+(\frac{1}{3})^{2}-\frac{1}{3}+1=\frac{1}{27}+\frac{1}{9}-\frac{1}{3}+1=\frac{1 + 3-9 + 27}{27}=\frac{22}{27}\).
When \(x=-1\), \(y=(-1)^{3}+(-1)^{2}-(-1)+1=-1 + 1+1 + 1=2\).

Answer:

A. \(y'(t)=-32t + 50\)
B. \(0\leq t<\frac{25}{16}\) seconds
C. \(t=\frac{25}{16}\) seconds
D. \(y=\frac{705}{16}=44.0625\) ft
For question 7: The points are \((\frac{1}{3},\frac{22}{27})\) and \((-1,2)\)