A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s = 80t - 16t^2. For what time interval is the ball less than 64 feet above the ground?
A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s = 80t - 16t^2. For what time interval is the ball less than 64 feet above the ground?
If returns back down at,
$\displaystyle 80t-16t^2=0$
$\displaystyle 16t(5-t)=0$
$\displaystyle t=0,5$
We need for $\displaystyle 0\leq t\leq 5$
To satisfy,
$\displaystyle 80t-16t^2 \leq 64 $
Thus,
$\displaystyle 16t^2-80t + 64 \geq 0$
$\displaystyle t^2 - 5t + 4 \geq 0$
$\displaystyle (t-1)(t-4) \geq 0$
We have two cases,
1)Both positive.
$\displaystyle t-1>0$
$\displaystyle t-4>0$
Thus,
$\displaystyle t>4$
2)Both negative,
$\displaystyle t-1<0$
$\displaystyle t-4<0$
Thus,
$\displaystyle t<1$.
Thus, the time intervals are:
$\displaystyle 0\leq t\leq 1 \mbox{ and } 4\leq t \leq 5$
Hello, symmetry!
I'll baby-step through the explanation . . .
A ball is thrown vertically upward with an initial velocity of 80 ft/sec.
The height $\displaystyle s$ (in feet) of the ball after $\displaystyle t$ seconds is: $\displaystyle s \:= \:80t - 16t^2$
For what time interval is the ball less than 64 feet above the ground?
When is the height of the ball exactly 64 feet?
We have: .$\displaystyle 80t - 16t^2 \:=\:60\quad\Rightarrow\quad t^2 - 5t + 4 \:=\:0$
. . $\displaystyle (t - 1)(t - 4)\:=\:0\quad\Rightarrow\quad t \:=\:1,\,4$
So the ball is takes 1 second to reach a height of 64 feet.
. . The time interval is: .$\displaystyle \boxed{[0,\,1)}$
Then it goes beyond 64 feet to its maximum height, then falls back down,
. . passing the 64-foot mark at $\displaystyle t = 4$.
So it is again below 64 feet from $\displaystyle t = 4$ until it hits the ground.
. . Uh-oh! .When does it hit the ground?
The ball is on the ground when $\displaystyle s = 0$.
So we have: .$\displaystyle 80t - 16t^2\:=\:0$
. . Then: .$\displaystyle 16t(5 - t)\:=\:0\quad\Rightarrow\quad t \:=\:0,\,5$
So the ball is on the ground when $\displaystyle t = 0$ (of course!)
. . and again when $\displaystyle t = 5$.
So it is below 64 feet for last second: .$\displaystyle \boxed{(4,\,5]}$