Question

a model rocket is launched with an initial upward velocity of 50 m/s. the rockets height h (in meters) after t seconds is given by the following.

$h = 50t - 5t^{2}$

find all values of t for which the rockets height is 20 meters.

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 \( h = 20 \) meters, and the height formula is \( h = 50t - 5t^{2} \). So we set up the equation:
\( 20 = 50t - 5t^{2} \)
Rearrange it to the standard quadratic form \( ax^{2}+bx + c = 0 \):
\( 5t^{2}-50t + 20 = 0 \)
We can simplify this equation by dividing each term by 5:
\( t^{2}-10t + 4 = 0 \)

Step2: Use the quadratic formula

For a quadratic equation \( ax^{2}+bx + c = 0 \), the quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \).
In our equation \( t^{2}-10t + 4 = 0 \), we have \( a = 1 \), \( b=- 10 \), and \( c = 4 \).
First, calculate the discriminant \( \Delta=b^{2}-4ac \):
\( \Delta=(-10)^{2}-4\times1\times4=100 - 16 = 84 \)
Then, find the values of \( t \):
\( t=\frac{-(-10)\pm\sqrt{84}}{2\times1}=\frac{10\pm\sqrt{84}}{2}=\frac{10\pm2\sqrt{21}}{2}=5\pm\sqrt{21} \)

Step3: Calculate the numerical values

Calculate \( \sqrt{21}\approx4.5837 \)
For the plus sign: \( t = 5+\sqrt{21}\approx5 + 4.5837=9.5837\approx9.58 \)
For the minus sign: \( t = 5-\sqrt{21}\approx5 - 4.5837 = 0.4163\approx0.42 \)

Answer:

\( t = 0.42 \) or \( t = 9.58 \)