Question

1. caleb hits a golf ball off a tee that is 1 meter above the ground. the path of the golf ball is modeled by

$g(x)=-4.9x^{2}+14.7x+1$
how many seconds after he hits the ball does it land back on the ground?
__________

Explanation:

Step1: Set the function equal to 0

When the ball hits the ground, \(g(x)=0\). So we set up the equation \(-4.9x^{2}+14.7x + 1=0\).

Step2: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a=-4.9\), \(b = 14.7\), \(c = 1\)), the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(14.7)^{2}-4\times(-4.9)\times1=216.09 + 19.6=235.69\).
Then, \(x=\frac{-14.7\pm\sqrt{235.69}}{2\times(-4.9)}=\frac{-14.7\pm15.35}{-9.8}\).
We get two solutions:
\(x_1=\frac{-14.7 + 15.35}{-9.8}=\frac{0.65}{-9.8}\approx - 0.07\) (rejected since time \(x\geq0\)).
\(x_2=\frac{-14.7-15.35}{-9.8}=\frac{-30.05}{-9.8}\approx3.07\).

Answer:

The golf - ball hits the ground approximately \(3.07\) seconds after it is hit.