Problem 49 - Fundamentals of Physics

Question

A hot-air balloon is ascending at the rate of \(12 \mathrm{~m} / \mathrm{s}\) and is \(80 \mathrm{~m}\) above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what spced docs it hit the ground?

Step-by-Step Solution

Answer

(a) The package takes approximately 5.27 seconds to reach the ground. (b) It hits the ground with a speed of approximately 39.3 m/s.

Step 1: Understand the initial conditions

The balloon ascends at \(12 \text{ m/s}\) and the package is initially \(80 \text{ m}\) above the ground when dropped. This means the package initially has an upward velocity of \(12 \text{ m/s}\) and is at an initial height \(s_0 = 80 \text{ m}\).

Step 2: Establish the equation of motion

The equation of motion for the height \(s(t)\) of the package as a function of time \(t\) can be expressed as \(s(t) = s_0 + v_0 t - \frac{1}{2} g t^2\), where \(v_0 = 12 \text{ m/s}\) is the initial velocity and \(g = 9.8 \text{ m/s}^2\) is the acceleration due to gravity.

Step 3: Set the final height condition

Since we want to find when the package hits the ground, set \(s(t) = 0\). This becomes the equation \(0 = 80 + 12t - \frac{1}{2} \times 9.8 t^2\).

Step 4: Solve the quadratic equation

Rearrange the equation to get \(-4.9t^2 + 12t + 80 = 0\). Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -4.9\), \(b = 12\), and \(c = 80\), to find \(t\).

Step 5: Calculate the time to reach ground

After applying the quadratic formula, the solutions for \(t\) are \(t = \frac{-12 \pm \sqrt{12^2 - 4(-4.9)(80)}}{2(-4.9)}\). Calculate this to find the positive value of \(t\) that satisfies the equation.

Step 6: Compute the speed upon impact

To find the speed at which it hits the ground, use the velocity equation \(v(t) = v_0 - gt\), where \(t\) is the time of impact found in Step 5. Substitute back \(v_0 = 12 \text{ m/s}\) and \(g = 9.8 \text{ m/s}^2\).

Step 7: Solve for the final velocity

Using the velocity equation \(v(t) = 12 - 9.8t\) and substituting the time \(t\) from Step 5, solve for \(v(t)\). Make sure the velocity is positive by focusing on its magnitude.

Key Concepts

Equations of Motion

Projectile motion problems often involve determining how an object moves under the influence of gravity. With equations of motion, we aim to understand more about how far, how fast, or how long an object travels when released at a certain height or velocity. These equations are powerful tools that can provide insight into the motion of objects.
For vertical motion like our package, we use an equation:

• \( s(t) = s_0 + v_0 t - \frac{1}{2} g t^2 \)

Here:

• \( s(t) \) represents the height at any time \( t \).
• \( s_0 \) is the initial height (80 meters in this problem).
• \( v_0 \) is the initial velocity (12 m/s upwards).
• \( g \) is the acceleration due to gravity (9.8 m/s²).

Using this equation helps to calculate when and how fast the package hits the ground by understanding its initial conditions and how it changes over time.
The goal is to set the height equal to zero to find when the package reaches the ground.

Quadratic Equation

The behavior of projectiles is often described using quadratic equations. This is because the path, or trajectory, of an object under the influence of gravity forms a parabolic shape. To solve for the time it takes for the package to hit the ground, we need to rearrange and solve the quadratic equation derived from our equation of motion.
The equation becomes:

• \( -4.9t^2 + 12t + 80 = 0 \)

This type of equation is solved using the quadratic formula:

• \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Where \( a = -4.9 \), \( b = 12 \), and \( c = 80 \). By calculating the discriminant \( b^2 - 4ac \), and subsequently finding the root values, we identify when the package will hit the ground. In projectile motion, both the positive value from this calculation is physically meaningful since it indicates the actual time duration of the fall.

Gravity Acceleration

Gravity is a key factor in projectile motion. It is the force that pulls objects downward towards the Earth, accelerating them at a constant rate of approximately 9.8 m/s². Understanding this constant acceleration is essential for solving projectile motion problems because it impacts both the vertical motion equations and the final velocity of the package when it reaches the ground.
When we calculate the package's velocity upon impact, the formula used is:

• \( v(t) = v_0 - gt \)

Where:

• \( v(t) \) is the velocity at time \( t \).
• \( v_0 \) is the initial velocity upwards.
• \( g \) is the gravity acceleration (9.8 m/s²).

Gravity works to slow down the upward motion before reversing it into a downward acceleration. By connecting time and velocity through these variables, you can solve for how fast the package will be moving as it impacts the ground, recognizing gravity's crucial role. Understanding and applying the constant acceleration due to gravity is what bridges initial conditions to the motion's final outcomes.