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Brut [27]
3 years ago
15

2.) Which ordered pairs, in the form (x, y) are solutions to the equation? A. Both are solutions. B. The first is not a solution

, but the second is. C. Neither is a solution. D. The first is a solution, but the second is not
Mathematics
1 answer:
marissa [1.9K]3 years ago
5 0

A. Both are solutions. is the answer

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3 years ago
Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.) A balloon, rising vert
Liula [17]

Answer:

(a) 2.79 seconds after its release the bag will strike the ground.

(b) At a velocity of 73.28 ft/second it will hit the ground.

Step-by-step explanation:

We are given that a balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 80 feet above the ground.

Assume the acceleration of the object is a(t) = −32 feet per second.

(a) For finding the time it will take the bag to strike the ground after its release, we will use the following formula;

                        s=ut+\frac{1}{2} at^{2}

Here, s = distance of the balloon above the ground = - 80 feet

          u = intital velocity = 16 feet per second

          a = acceleration of the object = -32 feet per second

          t = required time

So,  s=ut+\frac{1}{2} at^{2}

      -80=(16\times t)+(\frac{1}{2} \times -32 \times t^{2})

       -80=16t-16 t^{2}

       16 t^{2} -16t -80 =0

          t^{2} -t -5 =0

Now, we will use the quadratic D formula for finding the value of t, i.e;

           t = \frac{-b\pm \sqrt{D } }{2a}

Here, a = 1, b = -1, and c = -5

Also, D = b^{2} -4ac = (-1)^{2} -(4 \times 1 \times -5) = 21

So,  t = \frac{-(-1)\pm \sqrt{21 } }{2(1)}

      t = \frac{1\pm \sqrt{21 } }{2}

We will neglect the negative value of t as time can't be negative, so;

t = \frac{1+ \sqrt{21 } }{2} = 2.79 ≈ 3 seconds.

Hence, after 3 seconds of its release, the bag will strike the ground.

(b) For finding the velocity at which it hit the ground, we will use the formula;

                       v=u+at

Here, v = final velocity

So,  v=16+(-32 \times 2.79)

      v = 16 - 89.28 = -73.28 feet per second.

Hence, the bag will hit the ground at a velocity of -73.28 ft/second.

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Anyone know how to do this? Thanks!
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