Question

a model rocket is launched with an initial upward velocity of 188 ft/s. the rockets height h (in feet) after t seconds is given by the following.\\(h = 188t - 16t^2\\)\
find all values of t for which the rockets height is 92 feet.\
round your answer(s) to the nearest hundredth.\
(if there is more than one answer, use the \or\ button.)

Explanation:

Step1: Set up the equation

We know the height \( h = 92 \) feet, and the height formula is \( h = 188t - 16t^{2} \). So we set up the equation:
\( 188t - 16t^{2}=92 \)
Rearrange it to the standard quadratic form \( ax^{2}+bx + c = 0 \):
\( 16t^{2}-188t + 92 = 0 \)
We can simplify this equation by dividing all terms by 4:
\( 4t^{2}-47t + 23 = 0 \)

Step2: Use the quadratic formula

For a quadratic equation \( at^{2}+bt + c = 0 \), the quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \).
Here, \( a = 4 \), \( b=- 47 \), \( c = 23 \).
First, calculate the discriminant \( \Delta=b^{2}-4ac \):
\( \Delta=(-47)^{2}-4\times4\times23 \)
\( = 2209-368 \)
\( = 1841 \)
Then, find the values of \( t \):
\( t=\frac{47\pm\sqrt{1841}}{2\times4}=\frac{47\pm\sqrt{1841}}{8} \)

Step3: Calculate the two solutions

First solution:
\( t_{1}=\frac{47+\sqrt{1841}}{8}\approx\frac{47 + 42.9069}{8}=\frac{89.9069}{8}\approx11.24 \)
Second solution:
\( t_{2}=\frac{47-\sqrt{1841}}{8}\approx\frac{47 - 42.9069}{8}=\frac{4.0931}{8}\approx0.51 \)

Answer:

\( t = 0.51 \) or \( t = 11.24 \) seconds