Answer:
The time before the ballast hits the ground is 1.186918s
Explanation:
We assume that the upward direction is positive.
The bag released from the ballon is at rest. The velocity of the bag equals the velocity of the balloon.
The balloon has a velocity (V0y) = 3m/s
When the bag is released, it has a height of 5.36m
To touch the ground it makes a displacement of y = - 5.36m
The motion of the bag can be described as followed:
y = V0t + 1/2 * at²
t² + (2V0/a)*t - (2y/a) = 0
We can solve this equation for t:
t= ((-2V0/a) ± √((2V0/a)² + 4*(2y/a))) /2
t = -(V0/a) ± √((V0/a)² + (2y/a))
In this equation we can plug the given values ( V0 = 1.3 m/s ; y = 5.36m ; g =-9.8 m/s²)
t = (1.3/-9.8)±√((1.3/-9.8)²+(2*(-5.36)/-9.8))
t = 0.132653 ± 1.054266
t= 1.186918
The time before the ballast hits the ground is 1.186918s
