A hot-air balloon is rising straight up with a speed of 5.4 m/s. A ballast bag is released from rest relative to the balloon when it is 11.2 m above the ground. How much time elapses before the ballast bag hits the ground?
To find the time it takes for the ballast bag to hit the ground, we will use the kinematic equation:
v_f = v_i + at
Where:
v_f = final velocity (0 m/s, since the ballast bag hits the ground)
v_i = initial velocity (5.4 m/s, since the balloon is rising at this speed)
a = acceleration (acceleration due to gravity, which is -9.8 m/s^2)
t = time
First, let's find the time it takes for the ballast bag to reach the ground from rest relative to the balloon:
0 = 5.4 + (-9.8)t
Rearranging the equation, we have:
-9.8t = -5.4
Dividing both sides by -9.8, we get:
t = -5.4 / -9.8
Simplifying, t is approximately equal to 0.551 seconds.
However, we need to account for the initial height of the balloon. Since the ballast bag is released at a height of 11.2 m above the ground, it will take some additional time for the bag to fall that distance. We can use the equation:
s = ut + (1/2)at^2
Where:
s = distance (11.2 m)
u = initial velocity (0 m/s, since the bag is released from rest)
t = time
a = acceleration (-9.8 m/s^2)
Rearranging the equation, we have:
11.2 = 0 + (1/2)(-9.8)t^2
Simplifying, we get:
5.6t^2 = 11.2
Dividing both sides by 5.6, we have:
t^2 = 2
Taking the square root of both sides, we get:
t ā ā2
Approximating the square root of 2 to two decimal places, we have:
t ā 1.41 seconds
Therefore, the total time elapsed before the ballast bag hits the ground is approximately 0.551 seconds (time to fall from rest relative to the balloon) plus 1.41 seconds (time to fall the additional distance of 11.2 m). Adding these times together, the total time is approximately 1.96 seconds.