Question

a model rocket is launched with an initial upward velocity of 201 ft/s. the rocket’s height h (in feet) after t seconds is given by the following. ( h = 201t - 16t^2 ) find all values of ( t ) for which the rocket’s height is 93 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 formula \( h = 201t - 16t^{2} \) and we want to find \( t \) when \( h = 93 \). So we set up the equation:
\( 93 = 201t - 16t^{2} \)
Rearranging it to standard quadratic form \( ax^{2}+bx + c = 0 \), we get:
\( 16t^{2}-201t + 93 = 0 \)
Here, \( a = 16 \), \( b=- 201 \), \( c = 93 \)

Step2: Use the quadratic formula

The quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \)
First, calculate the discriminant \( D=b^{2}-4ac \)
\( D=(-201)^{2}-4\times16\times93 \)
\( D = 40401- 5952=34449 \)
Then, find the square root of the discriminant \( \sqrt{D}=\sqrt{34449} = 185.6 \) (wait, actually \( \sqrt{34449}=185.6\)? No, \( 185^{2}=34225 \), \( 186^{2}=34596 \), so my mistake, let's recalculate \( D \):

\( (-201)^{2}=201\times201 = 40401 \)

\( 4\times16\times93=64\times93 = 5952 \)

\( D=40401 - 5952=34449 \)

Wait, \( 185^{2}=34225 \), \( 186^{2}=34596 \), so \( \sqrt{34449}\approx185.60 \)? No, let's check \( 185.6^{2}=(185 + 0.6)^{2}=185^{2}+2\times185\times0.6+0.6^{2}=34225+222 + 0.36 = 34447.36 \), close to 34449. So \( \sqrt{34449}\approx185.61 \)

Now, \( t=\frac{201\pm\sqrt{34449}}{32} \)

First solution: \( t_{1}=\frac{201 + 185.61}{32}=\frac{386.61}{32}\approx12.08 \)

Second solution: \( t_{2}=\frac{201-185.61}{32}=\frac{15.39}{32}\approx0.48 \)

Answer:

\( t\approx0.48 \) seconds and \( t\approx12.08 \) seconds