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

Complete question :

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.

2.80seconds

Explanation:

Given that

Acceleration (a) = - 32

Height (s) at which sandbag was dropped = 80feet, therefore, distance (h) = - 80

Initial Velocity (Vo) = 16, which is the Velocity of the balloon at the time of releasing the sandbag.

Using the second equation of motion:

H = Vot + 1/2at^2

Where t= time taken

Inputting our values :

-80 = 16t + 0.5(-32)(t^2)

-80 = 16t - 16t^2

-16t^2 + 16t + 80 = 0

-t^2 + t + 5 = 0

Using the quadratic formula:

(-b±√b^2 - 4ac) / 2a

Where : a = - 1, b = 1 and c = 5

Using the quadratic equation solver:

[-1±√1^2 - 4(-1)(5)) / 2(-1)]

The roots are:

2.7912878474779

-1.7912878474779

Discard the negative root,

Therefore, time taken = 2.7912878474779 = 2.80s

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