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SSM An airplane with a speed of 97.5 m/s is climbing upward at an angle of 50.0 with respect to the horizontal. When the plane’s altitude is 732 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

Short Answer

Expert verified
(a) Distance is 1340 m. (b) The impact angle is -45.0 degrees.

Step by step solution

01

Define the problem and given data

We need to calculate two things: (a) the horizontal distance from the point directly beneath the release point of a package dropped from an airplane, and (b) the angle of the velocity vector of the package with respect to the ground just before it hits the earth. Given, the initial velocity of the plane is 97.5 m/s, the angle of climb is 50.0 degrees, and the altitude at release is 732 m.
02

Calculate horizontal and vertical components of initial velocity

The initial velocity of the package is the same as the plane's because it is released from it. Using trigonometry:- Horizontal component: \( v_{x_0} = v_0 \cos(\theta) = 97.5 \cos(50^\circ) \).- Vertical component: \( v_{y_0} = v_0 \sin(\theta) = 97.5 \sin(50^\circ) \).Substitute \( v_0 = 97.5 \) m/s and \( \theta = 50.0^\circ \) to calculate these components.
03

Calculate time of flight

To find the time \( t \) the package is in the air, use the vertical motion equation:\[ h = v_{y_0} t + \frac{1}{2}(-g)t^2 \]where the initial height \( h = 732 \) m, and \( g = 9.8 \) m/s² is the acceleration due to gravity (acting downwards). Solve this quadratic equation to find \( t \). This involves rearranging to form:\[ 0 = 732 + v_{y_0} t - \frac{1}{2} g t^2 \]
04

Calculate horizontal distance

Once the time \( t \) is determined, use it to find the horizontal distance \( x \) using the equation:\[ x = v_{x_0} \times t \]where \( v_{x_0} \) is the horizontal velocity component calculated earlier.
05

Calculate final vertical velocity

The final vertical velocity \( v_{y} \) can be found using:\[ v_{y} = v_{y_0} + (-g)t \]This gives the velocity in the vertical direction just before it hits the ground.
06

Calculate the angle relative to the ground

The angle \( \phi \) of the velocity vector just before impact is given by:\[ \tan(\phi) = \frac{v_{y}}{v_{x_0}} \]Calculate \( \phi \) using the arctan function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematics
Kinematics is a branch of physics that describes the motion of objects without considering the forces that cause this motion. To understand the problem of the package released from an airplane, we use the principles of kinematics to calculate its motion as it travels downwards. Kinematics involves various parameters such as velocity, acceleration, time, and displacement. In the problem, we make use of initial velocity, time of flight, and displacement to figure out the position and velocity of the package.
Kinematics equations help us relate these parameters. For example, the vertical motion equation \[ h = v_{y_0} t + \frac{1}{2}(-g)t^2 \]one of the main kinematic equations, allows us to determine the time the package is in the air.
  • Displacement refers to the change in position of the package.
  • Velocity is the speed in a given direction (both initial and final velocities are considered).
  • Acceleration due to gravity indicates how fast the velocity is changing vertically.
The combination of these concepts helps in predicting where the package will land in relation to the release point.
Trigonometry in Physics
Trigonometry is key in physics, especially when dealing with angles and dimensions that aren’t perfectly vertical or horizontal. In our exercise, trigonometry helps us break down the initial velocity of the airplane into two components: horizontal and vertical. By resolving these components using trigonometric functions, we can make more accurate calculations for the projectile motion.
The two main trigonometric equations used are:
  • \( v_{x_0} = v_0 \cos(\theta) \) describes the horizontal component of velocity.
  • \( v_{y_0} = v_0 \sin(\theta) \) describes the vertical component of velocity.

These components are critical because they each influence different parts of the motion. The horizontal component affects how far the package will travel along the ground, and the vertical component determines how long it stays in the air before hitting the ground. Understanding these concepts helps bridge the application of trigonometry from simple angle calculations to dynamic systems in motion.
Vertical and Horizontal Motion
Projectile motion, like the package dropped from the airplane, involves two main types of motion: vertical and horizontal, which occur simultaneously but independently. Understanding this separation is crucial in analyzing the problem.

Vertical Motion

Vertical motion is governed by gravity, which constantly accelerates the package downwards at \( g = 9.8 \) m/s². Initially, a vertical velocity component is present, making the package go upwards slightly before gravity pulls it back down. This motion follows the kinematic equation for time of flight, helping us establish how long the package will remain airborne.

Horizontal Motion

In contrast, horizontal motion is not accelerated since there’s no horizontal force acting on the package once released. Hence, the package moves horizontally at a constant speed, determined by the initial horizontal velocity component. The distance traveled horizontally is a straight-line path dependent on the time of flight and initial speed.
  • For horizontal distance, we use:\[ x = v_{x_0} \times t \]
  • The final position is purely influenced by the horizontal component.
Combining both motions, we can predict the trajectory and landing spot of the package accurately by calculating separately and then integrating the horizontal and vertical aspects of its motion.

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