Problem 60 - College Physics

Question

A helicopter 8.50 m above the ground and descending at \(3.50 \mathrm{~m} / \mathrm{s}\) drops a package from rest (relative to the helicopter). Just as it hits the ground, find (a) the velocity of the package relative to the helicopter and (b) the velocity of the helicopter relative to the package. The package falls freely.

Step-by-Step Solution

Answer

(a) 23.7 m/s downward, (b) -23.7 m/s upward.

Step 1: Define Known Values

First, identify the known values from the problem. The helicopter is descending at a constant velocity of \(3.50 \text{ m/s}\). It drops a package from a height of \(8.50 \text{ m}\). The package falls from rest, meaning its initial velocity relative to the ground is \(0 \text{ m/s}\), but since it falls with the helicopter's descent, its initial velocity is \(-3.50 \text{ m/s}\) relative to the helicopter's motion.

Step 2: Calculate the Time of Fall

Use the equation for free fall motion to determine the time it takes for the package to hit the ground. The vertical distance \( s = 8.50 \text{ m}\), and the initial velocity \( u = -3.50 \text{ m/s}\). Using the equation: \[ s = ut + \frac{1}{2}gt^2 \]Where \( g = 9.81 \text{ m/s}^2\). Substitute the values to solve for \( t \).

Step 3: Solve the Quadratic Equation

Substitute the known values in the equation: \[ 8.50 = (-3.50)t + \frac{1}{2}(9.81)t^2 \]Rearrange it to: \[ 4.905t^2 - 3.50t - 8.50 = 0 \]Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( t \). Here, \( a = 4.905, \ b = -3.50, \ c = -8.50 \).

Step 4: Calculate Velocity of Package Relative to Ground

Once \( t \) is found, calculate the velocity of the package just before hitting the ground using \[ v = u + gt \]Substitute \( u = -3.50 \text{ m/s}\) and \( g = 9.81 \text{ m/s}^2 \) into this equation.

Step 5: Determine Velocity Relative to Helicopter

Now find the velocity of the package relative to the helicopter. Use the relative velocity concept: \[ v_{rel} = v_{package} - v_{helicopter} \]The helicopter's velocity is the constant \(-3.50 \text{ m/s}\). Subtract this from the velocity of the package as determined in Step 4.

Step 6: Calculate Helicopter's Velocity Relative to Package

By definition, the velocity of the helicopter relative to the package is the negative of the velocity of the package relative to the helicopter. Thus, \[ v_{helicopter,rel} = -v_{rel} \]

Key Concepts

Free Fall Motion

In physics, free fall motion refers to the motion of an object under the influence of gravitational force only. When a package is dropped from a helicopter, like in this situation, it experiences free fall.
This means the only force acting on it is gravity, which in most cases on Earth can be represented as an acceleration of approximately 9.81 m/s² downward. This uniform acceleration allows us to predict the time it takes for the object to reach the ground using kinematic equations.
In free fall, despite starting with a certain initial velocity (as is the case with a package dropped from a moving helicopter), the object's velocity will increase linearly over time due to the effect of gravity. Understanding free fall is essential for solving many physics problems involving vertical motion.

Helicopter Descent

Helicopter descent is a scenario where the helicopter moves downwards at a constant velocity. In our original exercise, the helicopter descends at 3.50 m/s. This informs how we calculate the package's initial velocity: it shares the same downward motion at the moment it is released.
Helicopters can descend at various rates, often controllable by the pilot adjusting the rotors. The constant velocity descent makes calculations simpler, as acceleration in the helicopter's motion doesn’t need to be considered; only its constant velocity is relevant.
Recognizing the constant rate of descent allows us to determine the relative motion of the package concerning both the helicopter and the ground. This factor is critical in understanding the entire situation of the falling package.

Quadratic Equation in Physics

Quadratic equations frequently arise in physics, particularly when dealing with projectile motion. In our exercise, determining when the package hits the ground requires solving a quadratic equation.
The kinematic equation for vertical motion: \[ s = ut + \frac{1}{2}gt^2 \] leads to a quadratic format \[ 4.905t^2 - 3.50t - 8.50 = 0 \]. Solving this type of equation typically involves using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where the coefficients \(a\), \(b\), and \(c\) are identified from the equation.
In physics, solving quadratic equations is crucial as they describe the parabolic trajectory that most projectiles follow under uniform gravitational fields, excluding air resistance.

Initial Velocity Calculation

Initial velocity calculation in physics refers to determining an object's speed and direction at the start of a time period of interest. In this problem, as the package is released from a descending helicopter, its initial velocity relative to the ground is crucial.
The initial velocity is not zero because the package inherits the helicopter's speed, which is -3.50 m/s (indicating the downward direction). Setting this parameter allows us to analyze the entire motion from release until it reaches the ground.
Calculating initial velocity is fundamental when working with problems that involve motion, as it allows us to apply kinematic equations effectively to gauge how an object's velocity changes over time.