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 ≈ 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 ≈ -37.24 m/s
Therefore, the velocity of the ballast at impact was approximately -37.24 m/s (downward direction).
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?
1 answer
1 answer