A hot air balloon decends toward the ground with a velocity of (2.0 m/s)yhat. A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (6.0 m/s)xhat relative to the balloon. When opened, the bottle is 6.2 m above the ground. (Neglect air resistance.)
(a) What is the initial velocity of the cork, as seen by an observer on the ground? What is the speed of the cork, and its initial direction of motion, as seen by the same observer?
Speed 3 m/s
Direction 4° below the horizontal
etermine the maximum height above the ground attained by the cork.
How long does the cork remain in the air?
the initial velocity 6.3m/s
To find the initial velocity of the cork as seen by an observer on the ground, we need to consider the motion of the balloon and the cork separately.
The velocity of the balloon is given as (2.0 m/s)ŷ, which means it is descending with a constant velocity only in the y-direction. The cork is expelled horizontally from the balloon with a velocity of (6.0 m/s)x̂ relative to the balloon.
Since the cork is moving perpendicularly to the vertical motion of the balloon, it does not have any initial velocity in the y-direction relative to the ground. Hence, the initial velocity of the cork as seen by an observer on the ground is the same as its velocity relative to the balloon, which is (6.0 m/s)x̂.
To find the speed and initial direction of motion of the cork as seen by the ground observer, we can use vector addition.
The horizontal component of the speed of the cork is 6.0 m/s, and there is no vertical component of velocity. Therefore, the speed of the cork is calculated using the Pythagorean theorem as follows:
Speed of the cork = √((6.0 m/s)^2 + (0 m/s)^2) = 6.0 m/s
The initial direction of motion of the cork can be found by calculating the angle below the horizontal. The angle can be obtained using the inverse tangent function:
θ = tan^(-1)(0 m/s / 6.0 m/s) ≈ 0°
Thus, the initial direction of motion of the cork, as seen by the ground observer, is horizontal.
To determine the maximum height above the ground attained by the cork, we can use the equations of motion. The vertical motion of the cork is influenced by gravity only since we neglect air resistance.
The initial height of the cork is 6.2 m above the ground. The vertical acceleration due to gravity is -9.8 m/s^2 (taking negative sign as the upward direction). The final velocity when reaching the maximum height is 0 m/s since the cork momentarily comes to rest at the peak of its trajectory.
We can now use the kinematic equation:
v_f^2 = v_i^2 + 2aΔy
Since the final velocity is 0 m/s, we have:
0 = (6.0 m/s)^2 + 2(-9.8 m/s^2)(Δy)
Solving for Δy, we get:
Δy = (6.0 m/s)^2 / (2 * 9.8 m/s^2) ≈ 1.84 m
Therefore, the maximum height above the ground attained by the cork is approximately 1.84 m.
To determine how long the cork remains in the air, we can use the equation of motion to find the time it takes for the cork to reach its maximum height.
Using the equation:
v_f = v_i + at
Since the cork momentarily comes to rest at the maximum height, we have:
0 = (6.0 m/s) + (-9.8 m/s^2)t
Solving for t, we get:
t = (6.0 m/s) / (9.8 m/s^2) ≈ 0.61 s
Therefore, the cork remains in the air for approximately 0.61 seconds.