Problem 133 - Fundamentals of Physics

Question

A helicopter is flying in a straight line over a level field at a constant speed of \(6.20 \mathrm{~m} / \mathrm{s}\) and at a constant altitude of \(9.50 \mathrm{~m}\). A package is ejected horizontally from the helicopter with an initial velocity of \(12.0 \mathrm{~m} / \mathrm{s}\) relative to the helicopter and in a direction opposite the helicopter's motion. (a) Find the initial speed of the package relative to the ground. (b) What is the horizontal distance between the helicopter and the package at the instant the package strikes the ground? (c) What angle does the velocity vector of the package make with the ground at the instant before impact, as seen from the ground?

Step-by-Step Solution

Answer

(a) 5.80 m/s. (b) 16.68 m. (c) 66.33°.

Step 1: Determine Initial Speed of Package Relative to Ground

The package is ejected with a speed of \(12.0 \mathrm{~m/s}\) opposite to the helicopter's direction, which is moving at \(6.20 \mathrm{~m/s}\). To determine the package's speed relative to the ground, we subtract the helicopter's speed from the package's relative speed:\[ v_\text{package} = v_\text{helicopter} - v_\text{relative} = 6.20 \mathrm{~m/s} - 12.0 \mathrm{~m/s} = -5.80 \mathrm{~m/s} \]The negative sign indicates the package is moving in the opposite direction. The initial speed of the package is \(5.80 \mathrm{~m/s}\) relative to the ground.

Step 2: Calculate Time of Flight of Package

The package falls from a height of \(9.50 \mathrm{~m}\). To find the time it takes to hit the ground, use the formula for free fall: \[ y = \frac{1}{2}gt^2 \] where \( y = 9.50 \mathrm{~m}\) and \( g = 9.81 \mathrm{~m/s^2} \). Solving for \( t \):\[ 9.50 = \frac{1}{2} \times 9.81 \times t^2 \]\[ t^2 = \frac{9.50}{4.905} \approx 1.937 \]\[ t \approx \sqrt{1.937} \approx 1.39 \mathrm{~s} \]

Step 3: Calculate Horizontal Distance Between Helicopter and Package

With the time of flight known, we can determine the horizontal distance traveled by the package. Since the horizontal velocity is -5.80 m/s:\[ d = v_\text{horizontal} \times t = (-5.80 \mathrm{~m/s}) \times 1.39 \mathrm{~s} \approx -8.06 \mathrm{~m} \]The helicopter continues moving forward, covering: \[ d_\text{helicopter} = v_\text{helicopter} \times t = 6.20 \mathrm{~m/s} \times 1.39 \mathrm{~s} \approx 8.62 \mathrm{~m} \] The total distance between the helicopter and the package is the sum:\[ d_\text{total} = 8.62 + 8.06 = 16.68 \mathrm{~m} \].

Step 4: Determine Angle of Velocity Vector at Impact

The horizontal velocity remains \(-5.80 \mathrm{~m/s}\), and the vertical velocity component, at impact is calculated using:\[ v_y = gt = 9.81 \mathrm{~m/s^2} \times 1.39 \mathrm{~s} \approx 13.63 \mathrm{~m/s} \]The angle \( \theta \) with the horizontal is:\[ \tan(\theta) = \frac{v_y}{|v_x|} = \frac{13.63}{5.80} \approx 2.35 \]\[ \theta \approx \tan^{-1}(2.35) \approx 66.33^\circ \]

Key Concepts

Helicopter Physics

Understanding the movement of helicopters is crucial when studying projectile motion, especially when objects are dropped from or ejected off flying vehicles like helicopters. Helicopters fly by creating lift with their rotors, allowing them to hover, move vertically, and glide horizontally.
In our exercise, the helicopter moves horizontally at a constant speed of 6.20 m/s, maintaining an altitude of 9.50 meters. This consistent motion ensures that the helicopter is following a stable path parallel to the ground. When considering the ejection of an object, such as a package in this scenario, from a helicopter, we must incorporate both the speed and direction of the helicopter into our calculations. This allows us to determine how the object will behave relative to the stationary ground below.
By analyzing helicopter physics, we grasp how a constant speed and steady altitude impacts the motion of any objects released.

Free Fall

Free fall occurs when an object moves downward under the influence of gravity alone, with no air resistance considered. This concept is pivotal when calculating how long an object takes to hit the ground after being released from a height.
In the exercise example, the package initially at a height of 9.50 meters undergoes free fall immediately after ejection. The formula used to calculate its time in free fall is:
\[ y = \frac{1}{2}gt^2 \]Where:

• \(y\) is the vertical distance which is 9.50 meters
• \(g\) is the acceleration due to gravity, approximately 9.81 m/s²
• \(t\) represents the time in seconds

Solving for \(t\) gives us the time it takes for the package to reach the ground, which is a significant 1.39 seconds in this case.
Thus, free fall provides insights into how gravity impacts the object’s descent speed towards Earth.

Velocity Vector

The velocity vector is comprised of both magnitude and direction, representing the object’s overall speed and the path it's taking relative to a fixed point, like the ground.
For the package in the exercise, we explore its initial velocity as it's ejected from the helicopter. Its initial velocity components include:

• Horizontal: -5.80 m/s (opposite to the helicopter's forward motion)
• Vertical: Initially 0 m/s increasing due to gravitational acceleration

At any given moment, the total velocity can be visualized using these vector components in a two-dimensional plane. Upon impact, the package’s velocity vector is calculated, showing both horizontal and vertical components. The horizontal component remains constant at -5.80 m/s, while the vertical velocity increases due to gravity to approximately 13.63 m/s.
To find the angle \( \theta \) of impact relative to the ground, we use:\[ \tan(\theta) = \frac{v_y}{|v_x|} \]Where \( v_y \) is the vertical velocity, and \( v_x \) the horizontal velocity.
This understanding allows us to determine the angle of projection, which in this case is approximately 66.33 degrees.
Recognizing these components helps define the real-world trajectory and orientation of moving objects.

Horizontal Motion

Horizontal motion deals with how an object moves along the parallel axis of the Earth's surface, affected by the initial horizontal velocity and the time it takes for the object to descend due to gravity.
The package in our exercise is ejected horizontally at 12.0 m/s relative to the helicopter's motion. The resulting horizontal speed of the package relative to the ground is calculated to be -5.80 m/s. This negative sign indicates movement opposite to the helicopter.
The horizontal motion remains uniform because gravity only affects the vertical component. During the 1.39 seconds of free fall, the horizontal distance covered can be determined using:\[ d = v_{\text{horizontal}} \times t \]For the package:

• \(v_{\text{horizontal}} = -5.80\) m/s
• \(t = 1.39\) s

The package travels approximately -8.06 meters horizontally before striking the ground. Meanwhile, the helicopter continues its path, moving forward an additional 8.62 meters.
Understanding horizontal motion allows us to predict the exact landing spot of the package and its separation from the helicopter, accounting for both initial speeds and downward gravitational pull.