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?

Explanation:

Step1: Differentiate to find velocity

The derivative of $y(t)=-16t^{2}+50t + 5$ using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$ gives $y'(t)=\frac{d}{dt}(-16t^{2})+\frac{d}{dt}(50t)+\frac{d}{dt}(5)$. So $y'(t)=-32t + 50$.

Step2: Determine upward - motion interval

The ball is going upward when $y'(t)>0$. Set $y'(t)=-32t + 50>0$. Solving for $t$:
$-32t>-50$, then $t<\frac{50}{32}=\frac{25}{16}=1.5625$. Since $t\geq0$, the time interval is $0\leq t<\frac{25}{16}$ seconds.

Step3: Find time of maximum height

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

Step4: Calculate maximum height

Substitute $t = \frac{25}{16}$ into $y(t)$.
$y(\frac{25}{16})=-16(\frac{25}{16})^{2}+50(\frac{25}{16})+5$.
First, $-16(\frac{25}{16})^{2}=-16\times\frac{625}{256}=-\frac{625}{16}$.
Second, $50(\frac{25}{16})=\frac{1250}{16}$.
Then $y(\frac{25}{16})=-\frac{625}{16}+\frac{1250}{16}+5=\frac{-625 + 1250}{16}+5=\frac{625}{16}+5=\frac{625+80}{16}=\frac{705}{16}=44.0625$ feet.

Answer:

A. $y'(t)=-32t + 50$
B. $0\leq t<\frac{25}{16}$ seconds
C. $t=\frac{25}{16}=1.5625$ seconds
D. $44.0625$ feet