Q. 84 - Calculus

Question

Calvin uses a slingshot to launch an orange straight up in the air to see what will happen. The distance in feet be-tween the orange and the ground after t seconds is given by the equation $s\left(t\right)=-16t^2+90t+5$ Use this equation to answer the following questions:

1. What is the initial height of the orange? What is the initial velocity of the orange? What is the initial acceleration of the orange?
2. What is the maximum height of the orange?
3. When will the orange hit the ground?

Step-by-Step Solution

Answer

Part(a) The initial height, initial velocity  and initial acceleration of the orange is $5 feet$, $90$ feet per sec and $-30$ feet per square sec.

Part(b) The maximum height of the orange is $131$ feet . 

Part(c) The orange will hit the ground at $0.06\sec$.

Part(a) Step 1. Given Information

We are given a function,

$s\left(t\right)=-16t^2+90t+5$

Part(a) Step 2. Finding the initial height of the orange

The initial height of the orange is at $t=0$.

Therefore,

$$\begin{gathered} s\left(0\right)=-16·0^2+90·0+5 \\ s\left(0\right)=5 \end{gathered}$$

Hence, the initial height of the orange is $5$ feet.

Part(a) Step 3. Finding the initial velocity of the orange

The velocity is given by,

$$\begin{gathered} v\left(t\right)=\frac{d}{dt}s\left(t\right) \\ =\frac{d}{dt}\left(-16t^2+90t+5\right) \\ =-16·2t+90·1 \\ =-30 t+90 \end{gathered}$$

The initial velocity is at $t=0$. So,

$$\begin{gathered} v\left(0\right)=-30\times 0+90 \\ =90 \end{gathered}$$

Hence, the initial velocity of the orange is $90$feet per sec

Part(a) Step 4. Finding the initial acceleration of the orange

The acceleration is given by,

$$\begin{gathered} a\left(t\right)=\frac{d}{dt}v\left(t\right) \\ =\frac{d}{dt}(-30t+90) \\ =-30 \end{gathered}$$

Hence, the initial acceleration of the orange is $-30$ feet per square sec.

Part(b) Step 1. Finding the maximum height of the orange

The maximum height of the orange is given by,

$$\begin{gathered} s^{\text{'}}\left(t\right)=\frac{d}{dt}s\left(t\right) \\ =\frac{d}{dt}\left(-16t^2+90t+5\right) \\ =-16·2t+90·1 \\ =-30t+90 \end{gathered}$$

Substitute $s^{\text{'}}\left(t\right)=0$ to find the critical points,

$$\begin{gathered} s^{\text{'}}\left(t\right)=0 \\ -30 t+90=0 \\ 30t=90 \\ t=3 \end{gathered}$$

Calculate $s^{\text{'}\text{'}}\left(3\right)$ to find whether the height is maximum or minimum,

$$\begin{gathered} s^{\text{'}\text{'}}\left(t\right)=-30 \\ {s}^{\text{'}\text{'}}\left(3\right)=-30<0 \end{gathered}$$

Therefore, the maximum height of the orange is at $t=3$.

The maximum height of the orange is given by,

$$\begin{gathered} s\left(3\right)=-16·3^2+90·3+5 \\ =-144+270+5 \\ =131 \end{gathered}$$

Hence, the maximum height of the orange is $131$ feet .

Part(c) Step 1. Finding the time when the orange hit the ground.

The orange will hit the ground at $s\left(t\right)=0$, so

$$\begin{gathered} -16t^2+90t+5=0 \\ t=\frac{-90\pm \sqrt{{90}^2-4·(-16)5}}{2·(-16)} \\ =\frac{-90\pm 91.76}{30} \end{gathered}$$

Since, time never be negative, so

$$\begin{gathered} \frac{-90\pm 91.76}{30}=\frac{1.76}{30} \\ =0.06 \end{gathered}$$

Hence, the orange will hit the ground at $0.06\sec$.