Question

the equation $h(t) = -16t^2 + 89t$ 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 116 feet? round to two decimal places. select an answer: seconds, feet, seconds per foot, feet per second

Explanation:

Step1: Set up the equation

We know that \( h(t) = 116 \), so substitute into the height function:
\( -16t^{2}+89t = 116 \)
Rearrange to standard quadratic form \( ax^{2}+bx + c = 0 \):
\( 16t^{2}-89t + 116 = 0 \) (multiplying both sides by - 1)

Step2: Use the quadratic formula

The quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 16 \), \( b=-89 \), \( c = 116 \)
First, calculate the discriminant \( D=b^{2}-4ac \)
\( D=(-89)^{2}-4\times16\times116 \)
\( D = 7921-7424=497 \)

Then, find \( t \):
\( t=\frac{89\pm\sqrt{497}}{32} \)

Calculate the two solutions:
\( t_1=\frac{89 + \sqrt{497}}{32}\approx\frac{89 + 22.29}{32}=\frac{111.29}{32}\approx3.48 \)
\( t_2=\frac{89-\sqrt{497}}{32}\approx\frac{89 - 22.29}{32}=\frac{66.71}{32}\approx2.08 \)

Answer:

The rocket will be at a height of 116 feet at approximately \( t\approx2.08 \) seconds and \( t\approx3.48 \) seconds. (If we consider the context of the rocket's motion, both times are valid as the rocket goes up and then comes down. If we need to choose one, maybe the first time it reaches 116 feet is at \( t\approx2.08 \) seconds or the second time at \( t\approx3.48 \) seconds. Depending on the requirement, but from the calculation, the two solutions are approximately 2.08 and 3.48 seconds.)