A ballast bag is dropped from a balloon that is 280 m above the ground and rising at 14 m/s. What is the maximum height reached by the bag?
V = Vo + g*t = 0
14 - 9.8t = = 0.
9.8t = 14.
t = 1.43 s. To reach max h.
h = Vo*t + 0.5g*t^2=14*1.43 + 4.9*1.43^2
= 10 m.
To find the maximum height reached by the ballast bag, we can use the principle of conservation of energy. At the maximum height, all the initial potential energy of the bag is converted into kinetic energy.
First, let's determine the initial potential energy (PE initial) of the ballast bag when it is dropped from the balloon. The potential energy is given by the equation:
PE = mgh
where m is the mass of the bag (which we'll assume to be constant), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the initial height.
PE initial = mgh
Next, let's calculate the initial potential energy:
PE initial = m * 9.8 * 280
Now, let's determine the final kinetic energy (KE final) of the ballast bag at the maximum height. The kinetic energy is given by the equation:
KE = (1/2)mv^2
where m is the mass of the bag (which we'll assume to be constant), and v is the velocity when the bag reaches the maximum height.
KE final = (1/2)mv^2
Given that the balloon is rising at 14 m/s, the velocity of the bag (v) will also be 14 m/s at the moment it reaches the maximum height.
Now, let's calculate the final kinetic energy:
KE final = (1/2)m(14)^2
According to the principle of conservation of energy, the initial potential energy (PE initial) is equal to the final kinetic energy (KE final). Therefore:
PE initial = KE final
m * 9.8 * 280 = (1/2)m(14)^2
Let's simplify the equation by canceling out the mass of the bag (m):
9.8 * 280 = (1/2)(14)^2
Now, we can solve for the maximum height (h):
h = (1/2)(14)^2 / 9.8 * 280
h = (1/2)(196) / 9.8 * 280
h = 98 / 9.8 * 280
h = 10 * 280
h = 2800 m
Therefore, the maximum height reached by the ballast bag is 2800 meters.