15. A hot-air balloon is moving with a velocity of 2.1 m/s [up] when the balloonist
drops a ballast (a large mass used for height control) over the edge. The ballast
hits the ground 3.8 s later.
(a) How high was the balloon when the ballast was released?
(b) What was the velocity of the ballast at impact?
To solve this problem, we can use the equations of motion. Let's assume the initial position of the ballast when it was released is h and the final position when it hit the ground is 0.
(a) To find the initial height of the balloon, we can use the equation of motion:
h = h0 + v0t + (1/2)at^2
Since the balloon was moving up with a velocity of 2.1 m/s, we have:
h = h0 + 2.1t - (1/2)gt^2
The time taken for the ballast to hit the ground is 3.8 seconds, so we can substitute this value into our equation:
0 = h - (1/2)g(3.8)^2
Solving for h, we get:
h = (1/2)g(3.8)^2
Using the value of acceleration due to gravity g = 9.8 m/s^2, we can calculate the height:
h = (1/2)(9.8)(3.8)^2
h ā 69.62 m
Therefore, the balloon was approximately 69.62 meters high when the ballast was released.
(b) To find the velocity of the ballast at impact, we can use the equation:
v = v0 + at
In this case, the initial velocity is 0 m/s since the ballast was dropped, and the acceleration is due to gravity, -9.8 m/s^2. Substituting these values into the equation, we get:
v = 0 + (-9.8)(3.8)
v ā -37.24 m/s
Therefore, the velocity of the ballast at impact was approximately -37.24 m/s (downward direction).