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Chapter 4: Problem 33

A plane, diving with constant speed at an angle of \(53.0^{\circ}\) with the vertical, releases a projectile at an altitude of \(730 \mathrm{~m}\). The projectile hits the ground \(5.00 \mathrm{~s}\) after release. (a) What is the speed of the plane? (b) How far does the projectile travel horizontally during its flight? What are the (c) horizontal and (d) vertical components of its velocity just before striking the ground?

Short Answer

Expert verified
(a) Speed of plane: 26.53 m/s. (b) Horizontal distance: 79.59 m. (c) Horizontal velocity: 15.918 m/s. (d) Vertical velocity: 70.275 m/s.

Step by step solution

01

Determine Initial Vertical Position

Given that the projectile is released at an altitude of 730 m, we start solving the problem with this known height and the time it takes to strike the ground being 5.00 s.
02

Calculate Initial Vertical Velocity

Use the equations of motion to find the initial vertical velocity. We know that the vertical position can be described by the equation: \[ y = y_0 + v_{y_0}t - \frac{1}{2}gt^2 \]where \(y_0 = 730\) m, \(y = 0\) since it hits the ground, and \(g = 9.81\, \text{m/s}^2\). Since time \(t = 5\) s:\[ 0 = 730 + v_{y_0} \times 5 - \frac{1}{2} \times 9.81 \times 5^2 \]Solving for \(v_{y_0}\), we find that:\[ v_{y_0} = \frac{1}{2} \times 9.81 \times 25 - 730 \times \frac{1}{5} = 21.225 \, \text{m/s} \]
03

Calculate Initial Speed and Horizontal Velocity Component

The angle with the vertical is \(53.0^{\circ}\), which implies the angle with the horizontal is \(37.0^{\circ}\). From the vertical velocity formula: \[ v_{y_0} = v_0 \cdot \cos(37^{\circ}) \]We already found \(v_{y_0}\) is 21.225 m/s, hence:\[ 21.225 = v_0 \cdot \cos(37^{\circ}) \]\[ v_0 = \frac{21.225}{0.8} = 26.53 \text{ m/s} \]Now calculate the horizontal component: \[ v_{x_0} = v_0 \cdot \sin(37^{\circ}) = 26.53 \cdot 0.6 = 15.918 \text{ m/s} \]
04

Calculate Horizontal Distance

The horizontal distance travelled only depends on the initial horizontal velocity and time. Use \[ x = v_{x_0} \times t = 15.918 \times 5 = 79.59 \text{ meters} \]
05

Compute the Final Vertical Velocity

To find the vertical component of its velocity just before striking the ground, use \[ v_{y} = v_{y_0} + gt = 21.225 + 9.81 \times 5 = 70.275 \text{ m/s} \]

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause this motion. In projectile motion, which is a specific type of kinematics, the motion is analyzed based on the path taken and the time elapsed. This involves breaking the movement into its horizontal and vertical components.
When solving a projectile motion problem, it's crucial to understand kinematics principles so you can determine quantities like displacement, velocity, and time correctly.
This involves using equations of motion to find various parameters, such as the initial and final speeds of an object.
In the given problem, the projectile released by a plane in motion is observed for kinematic parameters over time.
This helps determine how far the projectile will travel horizontally and how fast it moves right before hitting the ground.
Vertical and Horizontal Components
In projectile motion, disentangling vertical and horizontal components is a fundamental step. These components allow us to separate the analysis of vertical motion from horizontal motion.
  • The vertical component of velocity is affected by gravity. Therefore, velocity changes due to acceleration from gravity.
  • The horizontal component remains constant since no external horizontal forces act after release.
The original exercise involves a plane diving with an angle of 53 degrees, meaning the angle with the horizontal is 37 degrees. By understanding these angles, you separate the initial velocity into horizontal and vertical components:
  • Vertical velocity component: Dependent on cosine of the horizontal angle due to its relation to the vertical axis (21.225 m/s in this case).

  • Horizontal velocity component: Dependent on sine of the horizontal angle due to its relation to the horizontal axis (15.918 m/s).
These components help calculate further motion details, critical in solving projectile problems effectively.
Equations of Motion
Equations of motion play a pivotal role in kinematic analysis. They help establish relationships between distance, velocity, and time. The main equations to remember are:
  • The equation for vertical displacement: \(y = y_0 + v_{y_0}t - \frac{1}{2}gt^2\)
  • The formula for horizontal distance: \( x = v_{x_0} imes t \).
By utilizing these equations, we can discern vertical and horizontal motion characteristics.
For instance, for the vertical component, initial vertical velocity and gravitational acceleration allow calculations of height and final velocity. For horizontal analysis, consistent initial horizontal velocity and time yield traveled distance.
Utilizing these equations enables solving complex problems step-by-step, allowing understanding of motion's inception to its conclusion.
Physics Problem Solving
Physics problem solving involves systematic steps that allow clear identification and resolution of complex questions. Breaking down a problem into smaller, manageable parts is essential.
  • First, understand the problem, identifying known values such as initial height, angles, times, and constants like gravitational force.
  • Then break the initial velocity into vertical and horizontal components using trigonometric functions.
  • Next, apply relevant equations of motion to find the desired unknowns like speed, horizontal distance, and final velocities.
Through thoughtful application of physics concepts, students develop a clearer understanding and enhanced problem-solving capability.
It鈥檚 important to visualize the scenario and use diagrams where possible, aiding in structuring thoughts and solutions.
This methodical approach ultimately leads to accurate results and a deeper comprehension of the mechanics involved.

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Most popular questions from this chapter

A moderate wind accelerates a pebble over a horizontal \(x y\) plane with a constant acceleration \(\vec{a}=\left(5.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(7.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). At time \(t=0\), the velocity is \((4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}\). What are the (a) magnitude and (b) angle of its velocity when it has been displaced by \(12.0 \mathrm{~m}\) parallel to the \(x\) axis?

A soccer ball is kicked from the ground with an initial speed of \(19.5 \mathrm{~m} / \mathrm{s}\) at an upward angle of \(45^{\circ} .\) A player \(55 \mathrm{~m}\) away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground?

A batter hits a pitched ball when the center of the ball is \(1.22 \mathrm{~m}\) above the ground. The ball leaves the bat at an angle of \(45^{\circ}\) with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of \(107 \mathrm{~m}\). (a) Does the ball clear a \(7.32\) -m-high fence that is \(97.5 \mathrm{~m}\) horizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center?

The fast French train known as the TGV (Train 脿 Grande Vitesse) has a scheduled average speed of \(216 \mathrm{~km} / \mathrm{h} .\) (a) If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is to be limited to \(0.050 \mathrm{~g}\), what is the smallest radius of curvature for the track that can be tolerated? (b) At what speed must the train go around a curve with a \(1.00 \mathrm{~km}\) radius to be at the acceleration limit?

\(A\) is located \(4.0 \mathrm{~km}\) north and \(2.5 \mathrm{~km}\) east of ship \(B\). Ship \(A\) has a velocity of \(22 \mathrm{~km} / \mathrm{h}\) toward the south, and ship \(B\) has a velocity of \(40 \mathrm{~km} / \mathrm{h}\) in a direction \(37^{\circ}\) north of east. (a) What is the velocity of \(A\) relative to \(B\) in unit-vector notation with \(\hat{\mathrm{i}}\) toward the east? (b) Write an expression (in terms of \(\hat{\mathrm{i}}\) and \(\hat{\mathrm{j}}\) ) for the position of \(A\) relative to \(B\) as a function of \(t\), where \(t=0\) when the ships are in the positions described above. (c) At what time is the separation between the ships least? (d) What is that least separation?
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