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Chapter 3: Problem 32

An airplane with a speed of \(97.5 \mathrm{~m} / \mathrm{s}\) is climbing upward at an angle of \(50.0^{\circ}\) with respect to the horizontal. When the plane's altitude is \(732 \mathrm{~m}\), 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) The package travels 1,354.32 meters horizontally. (b) The impact velocity angle is approximately -65.6掳.

Step by step solution

01

Break Down Initial Velocity into Components

First, find the initial horizontal and vertical components of the velocity of the airplane. The initial velocity is given as \( v_0 = 97.5 \ \mathrm{m/s} \) and the climb angle is \( \theta = 50.0^\circ \).The horizontal component is given by:\[ v_{0x} = v_0 \cos \theta = 97.5 \times \cos(50^\circ) \approx 62.7 \ \mathrm{m/s} \]The vertical component is:\[ v_{0y} = v_0 \sin \theta = 97.5 \times \sin(50^\circ) \approx 74.7 \ \mathrm{m/s} \]
02

Calculate Time of Flight

Using the vertical motion equation, find the time taken for the package to reach the ground. Initial vertical position is \( y_0 = 732 \ \mathrm{m} \) and the final vertical position \( y = 0 \), with acceleration due to gravity \( a = -9.81 \ \mathrm{m/s}^2 \).Use the equation:\[ y = y_0 + v_{0y}t + \frac{1}{2}at^2 \]Substitute the known values:\[ 0 = 732 + 74.7t - \frac{1}{2}(9.81)t^2 \]Solving the quadratic equation:\[ 0 = -4.905t^2 + 74.7t + 732 \]Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -4.905 \), \( b = 74.7 \), and \( c = 732 \), the positive root is approximately \( t \approx 21.6 \ \, \mathrm{s} \).
03

Determine Horizontal Distance Travelled

Calculate the horizontal distance traveled by the package using the horizontal velocity and the time of flight.\[ x = v_{0x} \cdot t \]\[ x = 62.7 \times 21.6 \approx 1,354.32 \ \mathrm{m} \]
04

Calculate Final Vertical Velocity at Impact

Find the vertical velocity of the package just before impact using:\[ v_{y} = v_{0y} + at \]\[ v_{y} = 74.7 - 9.81 \times 21.6 \approx -137.2 \ \mathrm{m/s} \]
05

Determine Angle of Velocity Vector Before Impact

After finding the final vertical and horizontal velocities, calculate the angle \( \phi \) of the velocity vector relative to the horizontal.Use the formula:\[ \tan \phi = \frac{v_y}{v_{0x}} \]\[ \tan \phi = \frac{-137.2}{62.7} \approx -2.187 \]Find the angle:\[ \phi = \arctan(-2.187) \approx -65.6^\circ \]This angle is measured below the horizontal axis.

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Key Concepts

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

Horizontal and Vertical Components
Understanding how an object moves in both horizontal and vertical directions is crucial in projectile motion. For the exercise involving an airplane, this involves breaking down the plane's velocity into two components: horizontal and vertical.
  • The horizontal component (\( v_{0x} \)) is calculated using the cosine of the angle of ascent. This tells us how fast the airplane moves across the ground, ignoring altitude changes. For instance, if an airplane is moving at \( 97.5 \ \mathrm{m/s} \) and climbs at \( 50.0^{\circ} \), its horizontal speed becomes \( 62.7 \ \mathrm{m/s} \)
  • The vertical component (\( v_{0y} \)) is determined using the sine of the angle. It describes how quickly the plane ascends. In the same example, this would be \( 74.7 \ \mathrm{m/s} \).
By understanding these components, we can foresee how objects, like the package dropped from the airplane, behave separately in horizontal and vertical motions. These components are foundational for further calculations like time of flight and impact angle.
Time of Flight
Time of flight is the duration a projectile stays in motion before hitting the ground. To calculate this, we consider only the vertical motion.
  • The fundamental equation here involves the initial and final vertical positions, initial vertical velocity, and gravitational acceleration: \( y = y_0 + v_{0y}t + \frac{1}{2}at^2 \).
  • Given the plane's release from \( 732 \ \mathrm{m} \) with a vertical velocity of \( 74.7 \ \mathrm{m/s} \), and knowing gravity鈥檚 effect as \(-9.81 \ \mathrm{m/s^2} \), solving this quadratic equation gives us the time of flight.
In the airplane scenario, solving the equation gives approximations leading to the time of about \( 21.6 \ \mathrm{s} \).During this time, both components act simultaneously, though only vertical motion determines how long the object stays airborne.
Angle of Impact
The angle of impact is how the projectile makes contact with the ground. This angle reflects the trajectory just before landing and is influenced by the horizontal and final vertical velocities.
  • Once the horizontal velocity (\( v_{0x} \ = 62.7\ \mathrm{m/s} \)) and final vertical velocity (\( v_y = -137.2 \ \mathrm{m/s} \)) are known, \( \tan \phi = \frac{v_y}{v_{0x}} \) helps in finding the angle.
  • Using this formula, one calculates the angle \( \phi \) to find that the impact angle below the horizontal is approximately \(-65.6^{\circ} \).
This angle reveals how steeply or shallowly the projectile hits. It's important to determine potential impact force and orientation, especially crucial in real-world applications like air drops.

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Most popular questions from this chapter

Two trees have perfectly straight trunks and are both growing perpendicular to the flat horizontal ground beneath them. The sides of the trunks that face each other are separated by \(1.3 \mathrm{~m}\). A frisky squirrel makes three jumps in rapid succession. First, he leaps from the foot of one tree to a spot that is \(1.0 \mathrm{~m}\) above the ground on the other tree. Then, he jumps back to the first tree, landing on it at a spot that is \(1.7 \mathrm{~m}\) above the ground. Finally, he leaps back to the other tree, now landing at a spot that is \(2.5 \mathrm{~m}\) above the ground. What is the magnitude of the squirrel's displacement?

Mario, a hockey player, is skating due south at a speed of \(7.0 \mathrm{~m} / \mathrm{s}\) relative to the ice. A teammate passes the puck to him. The puck has a speed of \(11.0 \mathrm{~m} / \mathrm{s}\) and is moving in a direction of \(22^{\circ}\) west of south, relative to the ice. What are the magnitude and direction (relative to due south) of the puck's velocity, as observed by Mario?

A swimmer, capable of swimming at a speed of \(1.4 \mathrm{~m} / \mathrm{s}\) in still water (i.e., the swimmer can swim with a speed of \(1.4 \mathrm{~m} / \mathrm{s}\) relative to the water), starts to swim directly across a 2.8-km- wide river. However, the current is \(0.91 \mathrm{~m} / \mathrm{s}\), and it carries the swimmer downstream, (a) How long does it take the swimmer to cross the river? (b) How far downstream will the swimmer be upon reaching the other side of the river?

Concept Simulation 3.2 at reviews the concepts that are important in this problem. A golfer imparts a speed of \(30.3 \mathrm{~m} / \mathrm{s}\) to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation. (a) How much time does the ball spend in the air? (b) What is the longest "hole in one" that the golfer can make, if the ball does not roll when it hits the green?

Two cars, \(A\) and \(B,\) are traveling in the same direction, although car \(A\) is \(186 \mathrm{~m}\) behind car \(\mathrm{B}\). The speed of \(\mathrm{A}\) is \(24.4 \mathrm{~m} / \mathrm{s},\) and the speed of \(\mathrm{B}\) is \(18.6 \mathrm{~m} / \mathrm{s}\). How much time does it take for A to catch B?
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