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;
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}](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7D%20at%5E%7B2%7D)
![-80=(16\times t)+(\frac{1}{2} \times -32 \times t^{2})](https://tex.z-dn.net/?f=-80%3D%2816%5Ctimes%20t%29%2B%28%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20-32%20%5Ctimes%20t%5E%7B2%7D%29)
![-80=16t-16 t^{2}](https://tex.z-dn.net/?f=-80%3D16t-16%20t%5E%7B2%7D)
![16 t^{2} -16t -80 =0](https://tex.z-dn.net/?f=16%20t%5E%7B2%7D%20-16t%20-80%20%3D0)
![t^{2} -t -5 =0](https://tex.z-dn.net/?f=t%5E%7B2%7D%20-t%20-5%20%3D0)
Now, we will use the quadratic D formula for finding the value of t, i.e;
![t = \frac{-b\pm \sqrt{D } }{2a}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-b%5Cpm%20%5Csqrt%7BD%20%7D%20%7D%7B2a%7D)
Here, a = 1, b = -1, and c = -5
Also, D =
=
= 21
So, ![t = \frac{-(-1)\pm \sqrt{21 } }{2(1)}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-%28-1%29%5Cpm%20%5Csqrt%7B21%20%7D%20%7D%7B2%281%29%7D)
![t = \frac{1\pm \sqrt{21 } }{2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B1%5Cpm%20%5Csqrt%7B21%20%7D%20%7D%7B2%7D)
We will neglect the negative value of t as time can't be negative, so;
= 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](https://tex.z-dn.net/?f=v%3Du%2Bat)
Here, v = final velocity
So, ![v=16+(-32 \times 2.79)](https://tex.z-dn.net/?f=v%3D16%2B%28-32%20%5Ctimes%202.79%29)
v = 16 - 89.28 = -73.28 feet per second.
Hence, the bag will hit the ground at a velocity of -73.28 ft/second.