Question

1. the function $y = -16t^2 + 25$ represents the height $y$ (in feet) of a pinecone $t$ seconds after falling from a tree. (section 8.2)

a. after how many seconds does the pinecone hit the ground?
b. a second pinecone falls from a height of 36 feet. which pinecone hits the ground in the least amount of time? explain.

Explanation:

Part (a)

Step1: Set \( y = 0 \) (ground level)

When the pinecone hits the ground, its height \( y = 0 \). So we set up the equation:
\( 0=-16t^{2}+25 \)

Step2: Solve for \( t \)

First, rearrange the equation:
\( 16t^{2}=25 \)
Then, divide both sides by 16:
\( t^{2}=\frac{25}{16} \)
Take the square root of both sides. Since time \( t\geq0 \), we consider the positive square root:
\( t = \sqrt{\frac{25}{16}}=\frac{5}{4} = 1.25\)

Part (b)

Step1: Find the function for the second pinecone

The height function for a falling object (assuming no air resistance) is \( y=-16t^{2}+h_{0} \), where \( h_{0} \) is the initial height. For the second pinecone, \( h_{0} = 36 \), so its height function is \( y=-16t^{2}+36 \).

Step2: Find when the second pinecone hits the ground

Set \( y = 0 \):
\( 0=-16t^{2}+36 \)
Rearrange:
\( 16t^{2}=36 \)
Divide by 16:
\( t^{2}=\frac{36}{16}=\frac{9}{4} \)
Take the positive square root (since \( t\geq0 \)):
\( t=\sqrt{\frac{9}{4}}=\frac{3}{2}=1.5 \)

Step3: Compare the times

The first pinecone hits the ground at \( t = 1.25 \) seconds, and the second at \( t = 1.5 \) seconds. Since \( 1.25<1.5 \), the first pinecone (from 25 feet) hits the ground in less time. This is because a lower initial height means the object has less distance to fall, so it takes less time (assuming the same acceleration due to gravity, modeled by the \( - 16t^{2}\) term).

Answer:

s:
a. \(\frac{5}{4}\) seconds (or 1.25 seconds)
b. The first pinecone (from 25 feet) hits the ground in the least amount of time because it takes 1.25 seconds, while the second takes 1.5 seconds.