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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3 years ago