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:
![y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}](https://tex.z-dn.net/?f=y%20%3D%20y_%7Bo%7D%20%2B%20v_%7Bo%7D%5Ccdot%20t%20%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20g%20%5Ccdot%20t%5E%7B2%7D)
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:
![120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}](https://tex.z-dn.net/?f=120%5C%2Cft%20%3D%200%5C%2Cft%20%2B%20%5Cleft%2895%5C%2C%5Cfrac%7Bft%7D%7Bs%7D%20%5Cright%29%5Ccdot%20t%20%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%28-32.174%5C%2C%5Cfrac%7Bft%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%20%5Ccdot%20t%5E%7B2%7D)
![-16.087\cdot t^{2} + 95\cdot t -120 =0](https://tex.z-dn.net/?f=-16.087%5Ccdot%20t%5E%7B2%7D%20%2B%2095%5Ccdot%20t%20-120%20%3D0)
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.