Question

a model rocket is launched with an initial upward velocity of 235 ft/s. the rockets height h (in feet) after t seconds is given by the following.
( h = 235t - 16t^2 )
find all values of t for which the rockets height is 151 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 that \( h = 235t - 16t^2 \) and we want to find \( t \) when \( h = 151 \). So we set up the equation:
\( 151 = 235t - 16t^2 \)
Rearranging this into standard quadratic form \( ax^2 + bx + c = 0 \), we get:
\( 16t^2 - 235t + 151 = 0 \)
Here, \( a = 16 \), \( b = -235 \), and \( c = 151 \).

Step2: Use the quadratic formula

The quadratic formula is \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
First, calculate the discriminant \( D = b^2 - 4ac \):
\( D = (-235)^2 - 4 \times 16 \times 151 \)
\( D = 55225 - 9664 \)
\( D = 45561 \)
Then, find the square root of the discriminant: \( \sqrt{45561} \approx 213.45 \) (we can check this by squaring 213.45: \( 213.45^2 \approx 45561 \))

Now, substitute into the quadratic formula:
\( t = \frac{235 \pm 213.45}{2 \times 16} \)

Step3: Calculate the two solutions

First solution (using the plus sign):
\( t_1 = \frac{235 + 213.45}{32} = \frac{448.45}{32} \approx 14.01 \)
Second solution (using the minus sign):
\( t_2 = \frac{235 - 213.45}{32} = \frac{21.55}{32} \approx 0.67 \)

Answer:

\( t \approx 0.67 \) or \( t \approx 14.01 \) seconds