Q. 80 - Calculus

Question

A bowling ball is thrown down from a $20$th-story window. After $3$ seconds, the bowling ball is $26$ feet from the ground and falling at a rate of $-106$ feet per second (downwards). You may assume that gravity causes a constant downward acceleration of $-32$ feet per second.

Part (a): If the height $s\left(t\right)$ of the bowling ball t seconds after being thrown is given by a quadratic polynomial function, use $s\left(3\right),s\text{'}\left(3\right),s\text{'}\text{'}\left(3\right)$ to find an equation for $s\left(t\right)$.

Part (b): How high is the $20$th-story window from which the bowling ball was thrown?

Part (c): How fast was the bowling ball initially thrown?

Step-by-Step Solution

Answer

Part (a): The equation of $s\left(t\right)$ is $s\left(t\right)=-16t^2-10t+200$.

Part (b): The height of the building is $200 ft$.

Part (c):  The initial velocity of the ball is $10 ft per second$.

Part (a) Step 1. Given information.

 Consider the given question,

$$\begin{gathered} s\left(3\right)=26, \\ s\text{'}\left(3\right)=-106, \\ s\text{'}\text{'}\left(3\right)=-32 \end{gathered}$$

Part (a) Step 2. Differentiate the function.

Consider the distance function,

$s\left(t\right)=a{t}^2+bt+c$

Differentiating the function,

$$\begin{gathered} s\text{'}\left(t\right)=2at+b, \\ s\text{'}\text{'}\left(t\right)=2a \end{gathered}$$

Using the condition to form three equations,

$$\begin{gathered} s\left(3\right)=a·3^2+b·3+c \\ 26=9a+3b+c ...... \left(i\right) \\ s\text{'}\left(3\right)=2a\left(3\right)+b \\ -106=6a+b ...... \left(ii\right) \\ s\text{'}\text{'}\left(3\right)=2a \\ -32=2a \\ a=-16 \end{gathered}$$

Part (a) Step 3. Find an equation of s t .

Substitute $a=-16$ in equation (ii),

$$\begin{gathered} -106=6\left(-16\right)+b \\ -106=-96+b \\ b=-10 \end{gathered}$$

Substitute the values of a, b in equation (i),

$$\begin{gathered} 26=9\left(-16\right)+3\left(-10\right)+c \\ 26=-144-30+c \\ c=200 \end{gathered}$$

Substitute the values of a, b, c in distance function,

$s\left(t\right)=-16t^2-10t+200$

Part (b) Step 1. Find the height of the building.

Consider the given question,

The moment when the ball is released from the top, $t=0$.

Substitute $t=0$ in the distance function,

$$\begin{gathered} s\left(0\right)=-16{\left(0\right)}^2-10\left(0\right)+200 \\ =200 \end{gathered}$$

Part (c) Step 1. Given information.

Consider the given question,

The initial velocity of the object when it was released, $t=0$.

Substitute $t=0$ and the values in $s\text{'}\left(0\right)$,

$$\begin{gathered} s\text{'}\left(0\right)=2\left(-16\right)\left(0\right)-10 \\ s\text{'}\left(0\right)=-10 \end{gathered}$$