Answer:
The time taken to reach the ground is t=2.146 approx 2 seconds.
Step-by-step explanation:
Given : An arrow is shot upward. Its height h, in feet, is given by the equation \(h(t)=-16t^2+32t+5\), where t is the time in seconds.
To find : How many seconds does it take until the arrow hits the ground?
Solution :
We have given the equation, \(h(t)=-16t^2+32t+5\)
where, h is the height in feet and t is the time in seconds.
We have to find in how many seconds does it take until the arrow hits the ground i.e. height became zero or h(t)=0
Substituting in the equation,
\(0=-16t^2+32t+5\)
Solve by quadratic formula,
Solution of equation \(ax^2+bx+c=0\) is \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
On comparing, a=-16 , b=32 , c=5
Solution is
\(t=\frac{-32\pm\sqrt{(32)^2-4(-16)(5)}}{2(-16)}\)
\(t=\frac{-32\pm\sqrt{1024+320}}{-32}\)
\(t=\frac{-32\pm\sqrt{1344}}{-32}\)
\(t=\frac{-32\pm 36.66}{-32}\)
\(t=\frac{-32+36.66}{-32},\frac{-32-36.66}{-32}\)
\(t=-0.146,2.146\)
We reject t=-0.146.
So, The time taken to reach the ground is t=2.146 approx 2 seconds.
