Question

h(x) = 5x² + 10x + 2 represents the height of a football in meters x seconds after it was thrown
a. write an equation that can be solved to determine when the football will reach the ground (at a height of 0 meters)
b. determine the number of seconds it will take for the football to reach the ground

Explanation:

Part A: Set up the equation

To find when the football is at 0 meters (ground level), we set the height function \( h(x) = 0 \). The height function is \( h(x)= -5x^{2}+10x + 2 \) (assuming the first term is \(-5x^{2}\) as a football's height should be a downward - opening parabola, maybe a typo in the original, but we'll proceed with the standard projectile motion quadratic). So the equation is \( - 5x^{2}+10x + 2=0 \).

Part B: Solve the quadratic equation

We use the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) for a quadratic equation \( ax^{2}+bx + c = 0 \). Here, \( a=-5 \), \( b = 10 \), \( c = 2 \).

Step 1: Calculate the discriminant \( D=b^{2}-4ac \)

Substitute the values: \( D=(10)^{2}-4\times(-5)\times2=100 + 40=140 \)

Step 2: Apply the quadratic formula

\( x=\frac{-10\pm\sqrt{140}}{2\times(-5)}=\frac{-10\pm2\sqrt{35}}{-10}=\frac{10\mp2\sqrt{35}}{10}=1\mp\frac{\sqrt{35}}{5} \)

We discard the negative solution for time (since time can't be negative in this context). \( \frac{\sqrt{35}}{5}\approx\frac{5.916}{5}\approx1.183 \), so \( x = 1+\frac{\sqrt{35}}{5}\approx1 + 1.183=2.183 \) seconds (or we can also consider the other root \( x = 1-\frac{\sqrt{35}}{5}\approx1 - 1.183=- 0.183 \) which we discard as time can't be negative).

Answer:

Part A: The equation is \(\boldsymbol{-5x^{2}+10x + 2 = 0}\)

Part B: The time is approximately \(\boldsymbol{2.18}\) seconds (or \(1+\frac{\sqrt{35}}{5}\) seconds)