Respuesta :
Given:
A tee box is 128 feet above its fairway. When a golf ball is hit from the tee box with an initial vertical velocity of 32 ft/s, the quadratic equation [tex]0=-16t^2+32t+128[/tex] gives the time in seconds when a golf ball is at height 0 feet on the fairway.
We need to determine the time that ball is in the air.
Time taken:
The time can be determined by factoring the quadratic equation.
Thus, we have;
[tex]-16t^2+32t+128=0[/tex]
Let us solve the equation using the quadratic formula,
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
Substituting [tex]a=-16, b=32, c=128[/tex] in the above formula, we get;
[tex]t=\frac{-32 \pm \sqrt{32^{2}-4(-16) 128}}{2(-16)}[/tex]
[tex]t=\frac{-32 \pm \sqrt{1024+8192}}{-32}[/tex]
[tex]t=\frac{-32 \pm \sqrt{9216}}{-32}[/tex]
[tex]t=\frac{-32 \pm 96}{-32}[/tex]
Thus, the roots of the equation are
[tex]t=\frac{-32+96}{-32}[/tex] and [tex]t=\frac{-32-96}{-32}[/tex]
[tex]t=\frac{64}{-32}[/tex] and [tex]t=\frac{-128}{-32}[/tex]
[tex]t=-2[/tex] and [tex]t=4[/tex]
Since, the value of t cannot be negative, thus, the value of t is [tex]t=4[/tex]
Hence, the ball is in the air for 4 seconds.
