Assume the acceleration of the object is a(t) = −32 feet per second per second. (Neglect air resistance.) 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.(a) How many seconds after its release will the bag strike the ground? (b) At what velocity will it hit the ground?

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;

[tex]s=ut+\frac{1}{2} at^{2}[/tex]

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, [tex]s=ut+\frac{1}{2} at^{2}[/tex]

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

[tex]-80=16t-16 t^{2}[/tex]

[tex]16 t^{2} -16t -80 =0[/tex]

[tex]t^{2} -t -5 =0[/tex]

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

[tex]t = \frac{-b\pm \sqrt{D } }{2a}[/tex]

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

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

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

[tex]t = \frac{1\pm \sqrt{21 } }{2}[/tex]

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

[tex]t = \frac{1+ \sqrt{21 } }{2}[/tex] = 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;

[tex]v=u+at[/tex]

Here, v = final velocity

So, [tex]v=16+(-32 \times 2.79)[/tex]

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

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