Understanding the Stone Dropped from a Balloon Problem

This problem involves relative motion and the equations of motion under constant acceleration (gravity). When the stone is dropped from the balloon, it initially has the same upward velocity as the balloon. After being dropped, the stone is only under the influence of gravity, which acts downwards.

Initial Conditions and Key Concepts

• The balloon is moving upwards with a uniform velocity (\(v_{balloon} = 5 \text{ m/s}\)).
• When the stone is dropped, its initial velocity relative to the ground is equal to the balloon's velocity at that moment, which is \(u = +5 \text{ m/s}\) (taking upward direction as positive).
• The height of the balloon from the ground when the stone is dropped is \(h_{initial} = 50 \text{ m}\).
• The acceleration acting on the stone after it is dropped is due to gravity, \(a = -g = -10 \text{ m/s}^2\) (since gravity acts downwards).
• The stone hits the ground, which means its total displacement from the point of dropping is \(\Delta y = -50 \text{ m}\) (since the ground is 50 m below the drop point).

Calculating the Time Taken for the Stone to Reach the Ground

We use the second equation of motion for the stone:

\(\Delta y = ut + \frac{1}{2}at^2\)

Substituting the known values:

\(-50 = (5)t + \frac{1}{2}(-10)t^2\)

\(-50 = 5t - 5t^2\)

Rearrange the equation into a standard quadratic form \(at^2 + bt + c = 0\):

\(5t^2 - 5t - 50 = 0\)

Divide the entire equation by 5 to simplify:

\(t^2 - t - 10 = 0\)

Now, we solve for \(t\) using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a=1\), \(b=-1\), \(c=-10\).

\(t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-10)}}{2(1)}\)

\(t = \frac{1 \pm \sqrt{1 + 40}}{2}\)

\(t = \frac{1 \pm \sqrt{41}}{2}\)

We get two possible values for \(t\). Since time must be positive, we take the positive root:

\(t = \frac{1 + \sqrt{41}}{2}\)

Using \(\sqrt{41} \approx 6.403\), the time is:

\(t \approx \frac{1 + 6.403}{2} = \frac{7.403}{2} \approx 3.7015 \text{ seconds}\)

Calculating the Distance the Balloon Travels

While the stone is falling, the balloon continues to move upwards with a uniform velocity of 5 m/s. The distance the balloon travels upwards during the time \(t\) the stone is in the air is given by:

\(Distance_{balloon} = Velocity_{balloon} \times Time\)

\(Distance_{balloon} = 5 \times \left(\frac{1 + \sqrt{41}}{2}\right)\)

\(Distance_{balloon} = \frac{5(1 + \sqrt{41})}{2} \text{ meters}\)

Using the approximate time \(t \approx 3.7015\) s:

\(Distance_{balloon} \approx 5 \times 3.7015 \approx 18.5075 \text{ meters}\)

Calculating the Final Height of the Balloon

The height of the balloon from the ground when the stone hits the ground is the initial height plus the distance the balloon travelled upwards during the stone's fall.

\(Height_{final} = Height_{initial} + Distance_{balloon}\)

\(Height_{final} = 50 + \frac{5(1 + \sqrt{41})}{2}\)

\(Height_{final} = \frac{100 + 5 + 5\sqrt{41}}{2} = \frac{105 + 5\sqrt{41}}{2}\)

Using the approximate value of \(\sqrt{41} \approx 6.403\):

\(Height_{final} \approx \frac{105 + 5 \times 6.403}{2} = \frac{105 + 32.015}{2} = \frac{137.015}{2} \approx 68.5075 \text{ meters}\)

Rounding to one decimal place, the height is approximately 68.5 m.

Conclusion

By calculating the time it takes for the stone to fall to the ground and the distance the balloon travels upwards during that same time, we find the final height of the balloon from the ground.