Question

unit 5 review
score: 30.5/71 answered: 7/18
question 8
the equation ( h(t) = -16t^2 + 69t ) models the height ( h ) in feet reached in ( t ) seconds by a model rocket propelled straight up from the ground. when will the rocket be at a height of 68 feet?
round to two decimal places.
select an answer
feet
seconds per foot
seconds
feet per second

Explanation:

Step1: Set up the equation

We know that \( h(t) = 68 \), so substitute into the given equation:
\( 68=-16t^{2}+69t \)
Rearrange it to standard quadratic form \( ax^{2}+bx + c = 0 \):
\( 16t^{2}-69t + 68 = 0 \)
Here, \( a = 16 \), \( b=-69 \), \( c = 68 \)

Step2: Use the quadratic formula

The quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \)
First, calculate the discriminant \( D=b^{2}-4ac \)
\( D=(-69)^{2}-4\times16\times68 \)
\( D = 4761-4352 \)
\( D = 409 \)

Then, find \( t \):
\( t=\frac{69\pm\sqrt{409}}{32} \)

Calculate the two solutions:
For the plus sign:
\( t_1=\frac{69 + \sqrt{409}}{32}\approx\frac{69+20.2237}{32}=\frac{89.2237}{32}\approx2.79 \)
For the minus sign:
\( t_2=\frac{69-\sqrt{409}}{32}\approx\frac{69 - 20.2237}{32}=\frac{48.7763}{32}\approx1.52 \)

Answer:

The rocket will be at a height of 68 feet at approximately \( t \approx 1.52 \) seconds and \( t \approx 2.79 \) seconds. (If we consider the context, both times are valid as the rocket goes up and then comes down. But if we need to present the answers as per the calculation, they are approximately 1.52 seconds and 2.79 seconds.)