Question

A bridge is 38 feet above a river. How many seconds does it take a rock dropped from the bridge to pass by a tree limb that is 10 feet above the water? (Use the model h(t)=h-16t)

Answer 1

SOLUTION

Given the question in the image, the following are the solution steps to get the number of seconds

STEP 1: Write the given model for h(t)

$h(t)=h_0-16t^2$

STEP 2: Write the given parameters

$\begin{gathered} h_0\text{ is the initial height of the bridge} \\ h(t)\text{ is the height of the rock after t seconds} \\ t=number\text{ of seconds} \\ \text{Hence,} \\ h_0=38,h(t)=10,t=\text{?} \end{gathered}$

STEP 3: Substitute the values into the given model to get t

$\begin{gathered} h(t)=h_0-16t^2 \\ By\text{ substitution,} \\ 10=38-16t^2 \\ By\text{ collecting like terms, we have;} \\ 10-38=-16t^2 \\ -28=-16t^2 \\ \text{Divide both sides by -16} \\ \frac{-28}{-16}=\frac{-16t^2}{-16} \\ t^2=\frac{28}{16}=\frac{7}{4}=1.75\Rightarrow\text{ Find the square root of both sides} \\ t=\sqrt[]{1.75}=1.322875656 \\ t\approx1.32\sec s \end{gathered}$

Hence, it will take the rock approximately 1.32 seconds to pass by the tree limb.