Question

1. if a ball is thrown vertically upward from the roof of 64 - foot building with a velocity of 64 ft/sec, its height after t seconds is s(t)=64 + 64t - 16t². a.) what is the maximum height the ball reaches? answer: 128 b.) what is the velocity of the ball when it hits the ground (height 0)? answer:

Explanation:

Step1: Find the time when the ball reaches maximum height

The height - function is $s(t)=64 + 64t-16t^{2}$. The velocity function $v(t)$ is the derivative of the height - function. Using the power rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we have $v(t)=s^\prime(t)=64-32t$. At the maximum - height, the velocity is 0. Set $v(t) = 0$:
$$64-32t = 0$$
Solve for $t$:
$$32t=64\Rightarrow t = 2$$

Step2: Find the maximum height

Substitute $t = 2$ into the height - function $s(t)$:
$$s(2)=64+64\times2-16\times2^{2}=64 + 128-64=128$$

Step3: Find the time when the ball hits the ground

Set $s(t)=0$, so $64 + 64t-16t^{2}=0$. Divide through by 16:
$$4 + 4t-t^{2}=0\Rightarrow t^{2}-4t - 4=0$$
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 1$, $b=-4$, and $c=-4$.
$$t=\frac{4\pm\sqrt{(-4)^{2}-4\times1\times(-4)}}{2\times1}=\frac{4\pm\sqrt{16 + 16}}{2}=\frac{4\pm\sqrt{32}}{2}=\frac{4\pm4\sqrt{2}}{2}=2\pm2\sqrt{2}$$
We take the positive value $t = 2 + 2\sqrt{2}$ since time cannot be negative.

Step4: Find the velocity when the ball hits the ground

Substitute $t = 2 + 2\sqrt{2}$ into the velocity function $v(t)=64-32t$:
$$v(2 + 2\sqrt{2})=64-32(2 + 2\sqrt{2})=64-64-64\sqrt{2}=-64\sqrt{2}$$

Answer:

a. 128
b. $-64\sqrt{2}$