I need help with these problems.
- If a bowman shoots an arrow straight up with an initial speed of 160 ft / sec, from a height of 8 feet, its height above the floor, in feet at time t expressed in seconds, is given by the function $h(t)=-16t^2+160t+8$.
a) What is the maximum height reached by the arrow?
$$t=\frac{-160}{2(-16)}=\frac{-160}{-32}=5$$
$$f(t)=f(5)=-16(5)^2+160(2)+8=408$$
The maximum height reached by the arrow is 408 feet.
b? How long does it take for the arrow to hit the ground?
$$-16t^2+160t+8=0$$ $$2t^2-20t-1=0$$
$$t=\frac{-(-20)\pm \sqrt{(-20)^2-4(2)(-1)}}{2(2)}$$
$$t=\frac{-(-20)\pm \sqrt{(-20)^2-4(2)(-1)}}{2(2)}$$
$$t=-0.05$$
$$t=10.05$$
Since the time can't be negative, the arrow hit the ground after 10.05 seconds.
c) How much is its horizontal travel (distance)?
My question is here. The horizontal distance is the absolute value of the difference between the $x$ values of theses points: (0,0) and (10.05,0)?
$$|10.05-0|=|10.05|=10.05$$
Does the distance is 10.05 feet? But I'm confused since the $x$ axis represents time.