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:
[tex]y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]
Where:
[tex]y_{o}[/tex] - Initial height of the ball, measured in feet.
[tex]v_{o}[/tex] - Initial speed of the ball, measured in feet per second.
[tex]g[/tex] - Gravitational constant, equal to [tex]-32.174\,\frac{ft}{s^{2}}[/tex].
[tex]t[/tex] - Time, measured in seconds.
Given that [tex]y_{o} = 0\,ft[/tex], [tex]v_{o} = 95\,\frac{ft}{s}[/tex], [tex]g = -32.174\,\frac{ft}{s^{2}}[/tex] and [tex]y = 120\,ft[/tex], the following second-order polynomial is found:
[tex]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}[/tex]
[tex]-16.087\cdot t^{2} + 95\cdot t -120 =0[/tex]
The roots of this polynomial are, respectively:
[tex]t_{1} \approx 4.075\,s[/tex] and [tex]t_{2} \approx 1.831\,s[/tex].
Both roots solutions are physically reasonable, since [tex]t_{1}[/tex] represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas [tex]t_{2}[/tex] 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.
