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?
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 (in feet) of the ball after seconds is:
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: .
. .
So the ball is takes 1 second to reach a height of 64 feet.
. . The time interval is: .
Then it goes beyond 64 feet to its maximum height, then falls back down,
. . passing the 64-foot mark at .
So it is again below 64 feet from until it hits the ground.
. . Uh-oh! .When does it hit the ground?
The ball is on the ground when .
So we have: .
. . Then: .
So the ball is on the ground when (of course!)
. . and again when .
So it is below 64 feet for last second: .