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, 




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

Here, a = 1, b = -1, and c = -5
Also, D =
=
= 21
So, 

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;

Here, v = final velocity
So, 
v = 16 - 89.28 = -73.28 feet per second.
Hence, the bag will hit the ground at a velocity of -73.28 ft/second.