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Leokris [45]
3 years ago
5

A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.

Mathematics
1 answer:
sattari [20]3 years ago
8 0

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}

Where:

y_{o} - Initial height of the ball, measured in feet.

v_{o} - Initial speed of the ball, measured in feet per second.

g - Gravitational constant, equal to -32.174\,\frac{ft}{s^{2}}.

t - Time, measured in seconds.

Given that y_{o} = 0\,ft, v_{o} = 95\,\frac{ft}{s}, g = -32.174\,\frac{ft}{s^{2}} and y = 120\,ft, 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}

-16.087\cdot t^{2} + 95\cdot t -120 =0

The roots of this polynomial are, respectively:

t_{1} \approx 4.075\,s and t_{2} \approx 1.831\,s.

Both roots solutions are physically reasonable, since t_{1} represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas t_{2} 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.

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