Answer:
The ball is at a height of 120 feet after 1.8 and 4.1 seconds.
Step-by-step explanation:
The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."
Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

Where:
- Initial height of the ball, measured in feet.
- Initial speed of the ball, measured in feet per second.
- Gravitational constant, equal to
.
- Time, measured in seconds.
Given that
,
,
and
, the following second-order polynomial is found:


The roots of this polynomial are, respectively:
and
.
Both roots solutions are physically reasonable, since
represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas
represents the instant when the ball the same height after reaching maximum height.
In nutshell, the ball is at a height of 120 feet after 1.8 and 4.1 seconds.