Question

if a football player passes a football from 4 feet off the groun initial velocity of 6 feet per second, how long will it take the football to hit the ground? use the equation h = -16t² and your answer to the nearest hundredth. 0.72 0.65 0.35 0.27

Explanation:

Step1: Substitute values into the equation

The height \( h = 0 \) (when it hits the ground), initial height \( h_0=4 \) feet and initial velocity \( v_0 = 6 \) feet per second. The equation \( h=-16t^{2}+v_0t + h_0 \) becomes \( 0=-16t^{2}+6t + 4 \).

Step2: Use the quadratic formula

For a quadratic equation \( ax^{2}+bx + c = 0 \) (\( a=-16\), \( b = 6\), \( c = 4 \)), the quadratic formula is \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \).
First, calculate the discriminant \( D=b^{2}-4ac=(6)^{2}-4\times(-16)\times4=36 + 256=292 \).
Then \( t=\frac{-6\pm\sqrt{292}}{2\times(-16)}=\frac{-6\pm2\sqrt{73}}{-32} \).
We have two solutions for \( t \):
\( t_1=\frac{-6 + \sqrt{292}}{-32}\) and \( t_2=\frac{-6-\sqrt{292}}{-32}\).
Since time \( t>0 \), we consider \( t=\frac{-6+\sqrt{292}}{-32}\approx\frac{-6 + 17.09}{-32}\) (rejected as it gives negative value) and \( t=\frac{-6-\sqrt{292}}{-32}=\frac{6+\sqrt{292}}{32}\approx\frac{6 + 17.09}{32}=\frac{23.09}{32}\approx0.72\).

Answer:

\(0.72\)