Question

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (\(\textbf{Fig. E2.44}\)). After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release does the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch \(a_y-t\), \(v_y-t\), and \(y-t\) graphs for the motion.

Short answer

The sandbag hits the ground after 3.29 s with a speed of 27.23 m/s. Its greatest height above ground is 41.27 m.

Step-by-step solution

Determine Initial Conditions

The initial velocity of the sandbag when released is the same as the balloon's, which is \( v_{0} = 5.00 \, \text{m/s} \) upward. The initial position \( y_{0} = 40.0 \, \text{m} \). After release, the only acceleration acting on the sandbag is due to gravity, which is \( a = -9.81 \, \text{m/s}^2 \) downward.

Position and Velocity at 0.250 s

To find the position and velocity at \( t = 0.250 \, \text{s} \), use the equations of motion:\[ y(t) = y_{0} + v_{0}t + \frac{1}{2}a t^{2} \]\[ v(t) = v_{0} + at \]Substitute \( t = 0.250 \, \text{s} \):Position: \[ y(0.250) = 40 + 5 \times 0.250 - \frac{1}{2} \times 9.81 \times (0.250)^2 = 41.23 \, \text{m} \]Velocity:\[ v(0.250) = 5 - 9.81 \times 0.250 = 2.55 \, \text{m/s} \]

Position and Velocity at 1.00 s

Substitute \( t = 1.00 \, \text{s} \) into the equations of motion:Position:\[ y(1.00) = 40 + 5 \times 1.00 - \frac{1}{2} \times 9.81 \times (1.00)^2 = 35.10 \, \text{m} \]Velocity:\[ v(1.00) = 5 - 9.81 \times 1.00 = -4.81 \, \text{m/s} \]

Time to Hit the Ground

Set \( y(t) = 0 \) and solve for \( t \):\[ 0 = 40 + 5t - \frac{1}{2} \times 9.81 \times t^2 \]Rearrange to form a quadratic equation:\[ 4.905t^2 - 5t - 40 = 0 \]Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):\[ t = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 4.905 \times (-40)}}{2 \times 4.905} \]Calculate:\[ t = 3.29 \, \text{s} \]

Impact Velocity

Calculate the velocity at the time of impact \( t = 3.29 \, \text{s} \):\[ v(3.29) = 5 - 9.81 \times 3.29 = -27.23 \, \text{m/s} \]

Greatest Height Above Ground

The sandbag reaches its maximum height when the velocity is zero.Set \( v(t) = 0 \):\[ 0 = 5 - 9.81t \]\[ t = \frac{5}{9.81} = 0.509 \, \text{s} \]Calculate the position at this time:\[ y(0.509) = 40 + 5 \times 0.509 - \frac{1}{2} \times 9.81 \times (0.509)^2 = 41.27 \, \text{m} \]

Sketch Graphs

For the graphs:- The acceleration \( a_y-t \) is constant at \(-9.81 \, \text{m/s}^2\).- The velocity \( v_y-t \) graph is a line with an initial value of \(5 \, \text{m/s}\) and a negative slope equal to \(-9.81\).- The position \( y-t \) graph is a downward-opening parabola starting from \(40 \, \text{m}\).

Key concepts

Kinematic Equations

Kinematic equations are a set of formulas that describe the motion of objects under uniform acceleration. They help us predict the future position and velocity of an object if we know its initial conditions. In the case of the sandbag, we use two primary kinematic equations:

1. Position equation: \[ y(t) = y_0 + v_0 t + \frac{1}{2} a t^2 \] This equation tells us how the position changes over time, given the initial position \( y_0 \), initial velocity \( v_0 \), and constant acceleration \( a \).

2. Velocity equation: \[ v(t) = v_0 + at \] This equation helps us find the velocity of the object at any time \( t \).

In projectile motion, like the sandbag falling, these equations are crucial to determine how high or far the object will travel, and its speed at different times. These calculations were used to find the position and velocity at both 0.250 s and 1.00 s after the release of the sandbag.

Free Fall

When an object is in free fall, it is moving under the influence of gravity alone, without any other forces acting on it. This state is ideal for studying the principles of projectile motion since it simplifies the calculations to just considering gravitational effects.

In our exercise, after the sandbag is released from the hot-air balloon, it enters into free fall. This means that the initial upward velocity of the sandbag (\( 5.00 \, \text{m/s} \)) is gradually reduced by frictionless gravitational pull until it comes to a momentary halt before falling back down to the ground.

The key point to remember in free fall is that air resistance is often negligible in physics problems. The only force at play here is gravity pulling the sandbag downwards.

Acceleration Due to Gravity

Acceleration due to gravity is a constant force acting on any object close to the Earth's surface, typically denoted as \( g \). Its value is approximately \( 9.81 \, \text{m/s}^2 \), directed downward. This acceleration plays a crucial role in both the equations of motion and the behavior of objects in free fall.

In the sandbag scenario, gravity not only affects the speed but also the trajectory path downward. During all calculations, the acceleration was assigned a negative sign to indicate the direction was opposite to the positive initial velocity of 5.00 m/s upward.

Understanding this concept helps in predicting when the sandbag stops rising and starts to fall, as gravity decelerates it upward before accelerating it downward during the fall.

Motion Graphs

In motion analysis, graphs are incredibly useful for visualizing changes in motion over time. They provide a clear representation of how position, velocity, and acceleration vary. Let's break down the three types of motion graphs sketched for this problem:

- **Acceleration vs. Time (\(a_y-t\))**: This graph is a straight line at \(-9.81 \, \text{m/s}^2\), indicating that gravity's acceleration is constant throughout the flight of the sandbag.

- **Velocity vs. Time (\(v_y-t\))**: This graph starts at \(5.00 \, \text{m/s}\), representing the initial upward velocity of the sandbag. The line has a constant negative slope because of the gravitational pull causing acceleration downwards. The velocity line crosses zero when the upward look stops momentarily, and continues as negative values, indicating the sandbag is moving downwards.

- **Position vs. Time (\(y-t\))**: This graph is a parabola, demonstrating the path of the sandbag. It starts at 40 m and rises slightly before peaking, depicting the highest point it reaches - after which, the path turns downwards showing descent until it hits the ground.