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.
Answer:
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
