Question

If a ball is thrown upward at 6464 feet per second from a height of 1212 ​feet, the height of the ball can be modeled by Sequals=1212plus+6464tminus−16t squaredt2 ​feet, where t is the number of seconds after the ball is thrown. How long after the ball is thrown is the height 7272 ​feet?
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Part 1
It takes enter your response here seconds for the ball to reach the height 72 feet.
​(Type an integer or a simplified fraction. Use a comma to separate answers as​ needed.) fractions

Answer 1

To determine how long after the ball is thrown it takes to reach a height of 7272 feet, we can set the given height equation equal to 7272 feet and solve for \( t \):

The height equation is:

\[
S(t) = 1212 + 6464t - 16t^2
\]

We need to set this equal to 7272:

\[
1212 + 6464t - 16t^2 = 7272
\]

First, we will simplify this equation by moving all terms to one side:

\[
1212 + 6464t - 16t^2 - 7272 = 0
\]

Combining like terms, we get:

\[
-16t^2 + 6464t - 6060 = 0
\]

To make the equation easier to work with, we'll multiply the equation by -1:

\[
16t^2 - 6464t + 6060 = 0
\]

This is a quadratic equation in standard form \( at^2 + bt + c = 0 \) where:
- \( a = 16 \)
- \( b = -6464 \)
- \( c = 6060 \)

Next, we can use the quadratic formula to solve for \( t \):

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Calculating the discriminant:

\[
b^2 - 4ac = (-6464)^2 - 4(16)(6060)
\]

Calculating \( (-6464)^2 \):

\[
6464^2 = 41704576
\]

Calculating \( 4 \times 16 \times 6060 \):

\[
4 \times 16 \times 6060 = 387840
\]

Now, subtract \( 387840 \) from \( 41704576 \):

\[
41704576 - 387840 = 41316736
\]

Now, substitute back into the quadratic formula:

\[
t = \frac{6464 \pm \sqrt{41316736}}{32}
\]

Calculating \( \sqrt{41316736} \):

\[
\sqrt{41316736} \approx 6426.43
\]

Now substituting this value back:

\[
t = \frac{6464 \pm 6426.43}{32}
\]

Calculating the two possible values for \( t \):

1. For the positive root:
\[
t = \frac{6464 + 6426.43}{32} \approx \frac{12890.43}{32} \approx 402.83
\]

2. For the negative root:
\[
t = \frac{6464 - 6426.43}{32} \approx \frac{37.57}{32} \approx 1.17
\]

Thus, the ball reaches a height of 7272 feet at approximately **1.17 seconds** and again at **402.83 seconds** after being thrown.

So, your final answer is:

\[
\text{It takes approximately } 1.17 \text{ seconds and } 402.83 \text{ seconds for the ball to reach 7272 feet.}
\]

In summary, the time it takes for the ball to reach the height of 7272 feet is approximately:
\[
1.17, 402.83
\]