/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 105 A projectile is launched with an... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A projectile is launched with an initial speed of \(30 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) above the horizontal. What are the (a) magnitude and (b) angle of its velocity \(2.0 \mathrm{~s}\) after launch, and \((\mathrm{c})\) is the angle above or below the horizontal? What are the (d) magnitude and (e) angle of its velocity \(5.0 \mathrm{~s}\) after launch, and \((\mathrm{f})\) is the angle above or below the horizontal?

Short Answer

Expert verified
At 2.0 s, velocity: 16.31 m/s at 23.24° above horizontal. At 5.0 s, velocity: 27.47 m/s at 57.35° below horizontal.

Step by step solution

01

Break Down Initial Velocity

Given that the initial speed of the projectile is 30 m/s and the launch angle is 60 degrees, we need to decompose this velocity into horizontal and vertical components.- The horizontal component of the velocity is given by \(v_{0x} = v_0 \cdot \cos(60^{\circ})\).- The vertical component of the velocity is given by \(v_{0y} = v_0 \cdot \sin(60^{\circ})\).Substituting the given values:\[v_{0x} = 30 \cdot \cos(60^{\circ}) = 30 \cdot \frac{1}{2} = 15 \, \text{m/s}\,\]\[v_{0y} = 30 \cdot \sin(60^{\circ}) = 30 \cdot \frac{\sqrt{3}}{2} = 15\sqrt{3} \, \text{m/s}\,\]
02

Determine Velocity Components at 2.0 s

To find the velocity of the projectile 2.0 s after launch, we calculate its horizontal and vertical components at this time.- The horizontal component \(v_x\) remains constant as \(v_{0x} = 15 \, \text{m/s}\).- The vertical component changes due to acceleration due to gravity. It's given by \(v_y = v_{0y} - g \cdot t\). Using \(g = 9.8 \, \text{m/s}^2\):\[v_y = 15\sqrt{3} - 9.8 \cdot 2 = 25.98 - 19.6 = 6.38 \, \text{m/s}\,\]
03

Find Magnitude and Angle at 2.0 s

Using the components from Step 2, calculate the magnitude \(v\) and direction \(\theta\) of the velocity:- The magnitude is given by \(v = \sqrt{v_x^2 + v_y^2}\):\[v = \sqrt{15^2 + 6.38^2} = \sqrt{225 + 40.7044} = \sqrt{265.7044} \approx 16.31 \, \text{m/s}\,\]- The angle with the horizontal is \(\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)\):\[\theta = \tan^{-1}\left(\frac{6.38}{15}\right) \approx 23.24^{\circ}\, \text{above the horizontal}.\]
04

Determine Velocity Components at 5.0 s

Now, calculate the velocity components at 5.0 s:- The horizontal component \(v_x\) is still 15 m/s.- The vertical component \(v_y\) is now:\[v_y = 15\sqrt{3} - 9.8 \cdot 5 = 25.98 - 49 = -23.02 \, \text{m/s}\,\] indicating it is moving downwards.
05

Find Magnitude and Angle at 5.0 s

Calculate the magnitude and direction of the velocity at 5.0 s using the results from Step 4:- The magnitude is:\[v = \sqrt{15^2 + (-23.02)^2} = \sqrt{225 + 529.8804} = \sqrt{754.8804} \approx 27.47 \, \text{m/s}\,\]- The angle with the horizontal is:\[\theta = \tan^{-1}\left(\frac{23.02}{15}\right) \approx 57.35^{\circ}\, \text{below the horizontal}.\]
06

Answer Questions

Based on the calculations: - (a) Magnitude of velocity at 2.0 s: approx. 16.31 m/s - (b) Angle at 2.0 s: approx. 23.24° above the horizontal - (c) The angle is above the horizontal. - (d) Magnitude of velocity at 5.0 s: approx. 27.47 m/s - (e) Angle at 5.0 s: approx. 57.35° - (f) The angle is below the horizontal.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Initial Velocity Components
When a projectile is launched, its initial velocity can be divided into two components: horizontal and vertical. This division is crucial for understanding the behavior of the projectile throughout its motion. By breaking down the initial velocity, we essentially translate a complex diagonal motion into more manageable horizontal and vertical motions.
  • Horizontal Component: This is found using the cosine of the launch angle. For a projectile launched at an angle \( \theta \), the horizontal component of the velocity \( v_{0x} \) is calculated as \( v_{0x} = v_0 \cdot \cos(\theta) \). In our example, with an initial speed \( v_0 = 30 \; \text{m/s} \) and a launch angle of \( 60^{\circ} \), the formula becomes \( v_{0x} = 30 \cdot \cos(60^{\circ}) = 15 \; \text{m/s} \).
  • Vertical Component: This is calculated using the sine of the launch angle. It is given by \( v_{0y} = v_0 \cdot \sin(\theta) \). For the given problem, \( v_{0y} = 30 \times \sin(60^{\circ}) = 15\sqrt{3} \; \text{m/s} \).
Understanding these components is essential because they remain independent throughout the projectile's flight, with gravity only affecting the vertical component.
Velocity Magnitude and Direction
Determining the magnitude and direction of a projectile's velocity at any point in time is key to analyzing its motion. After launch, the velocity isn’t just the initial speed; it changes as the projectile moves.

Calculating Magnitude

The magnitude of velocity is derived from the combined effect of horizontal and vertical velocity components. It is computed using the Pythagorean theorem: \( v = \sqrt{v_x^2 + v_y^2} \). For instance, after 2.0 seconds, with \( v_x = 15 \; \text{m/s} \) and \( v_y = 6.38 \; \text{m/s} \), the magnitude is \( \sqrt{15^2 + 6.38^2} \approx 16.31 \; \text{m/s} \).

Determining Direction

The direction at a given time \( \theta \) is the angle made with the horizontal line. It is found using the inverse tangent function: \( \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) \). With our components, after 2.0 seconds, the angle is \( \tan^{-1}\left(\frac{6.38}{15}\right) \approx 23.24^{\circ} \), which is above the horizontal.
Tracking these changes in magnitude and direction helps in predicting the projectile's path and where it will land.
Effect of Gravity on Projectile
Gravity plays a pivotal role in projectile motion by affecting the vertical component of the velocity. Unlike the horizontal component, which remains constant in the absence of air resistance, gravity continuously acts downwards on the projectile.
  • Constant Acceleration: Gravity imparts a constant acceleration of \( 9.8 \; \text{m/s}^2 \) downwards. This means the vertical velocity component decreases on the way up, becomes zero at the peak, and then increases in the downward direction.
  • Vertical Velocity Change: The formula used to determine the vertical component of the velocity at any time \( t \) is \( v_y = v_{0y} - g \cdot t \). For example, after 5 seconds, given \( v_{0y} = 15\sqrt{3} \; \text{m/s} \), the vertical velocity is \( 15\sqrt{3} - 9.8 \cdot 5 = -23.02 \; \text{m/s} \), indicating it's moving downwards.
Recognizing how gravity affects the projectile helps in understanding its trajectory, especially how high it will go and how long it will stay in the air.
Horizontal and Vertical Components
The separation of motion into horizontal and vertical components simplifies the analysis of projectile motion. These components behave independently, which allows us to solve problems more efficiently.
  • Horizontal Motion: Because there is no acceleration acting horizontally (ignoring air resistance), the horizontal component of velocity \( v_x \) remains constant throughout the flight. In this example, \( v_x = 15 \; \text{m/s} \) continues unchanged at both 2.0 seconds and 5.0 seconds.
  • Vertical Motion: Affected by gravity, this component changes. At 2.0 seconds, the vertical velocity effects are computed as previously shown. By 5.0 seconds, the downward influence of gravity results in a negative vertical velocity \( v_y = -23.02 \; \text{m/s} \).
By understanding these components, students can predict different stages of the projectile's motion, and how both horizontal and vertical influences guide its journey through the space.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A woman can row a boat at \(6.40 \mathrm{~km} / \mathrm{h}\) in still water. (a) If she is crossing a river where the current is \(3.20 \mathrm{~km} / \mathrm{h}\), in what direction must her boat be headed if she wants to reach a point directly opposite her starting point? (b) If the river is \(6.40 \mathrm{~km}\) wide, how long will she take to cross the river? (c) Suppose that instead of crossing the river she rows \(3.20 \mathrm{~km}\) down the river and then back to her starting point. How long will she take? (d) How long will she take to row \(3.20 \mathrm{~km} u p\) the river and then back to her starting point? (e) In what direction should she head the boat if she wants to cross in the shortest possible time, and what is that time?

An astronaut is rotated in a horizontal centrifuge at a radius of \(5.0 \mathrm{~m}\). (a) What is the astronaut's speed if the centripetal acceleration has a magnitude of \(7.0 \mathrm{~g} ?\) (b) How many revolutions per minute are required to produce this acceleration? (c) What is the period of the motion?

The position vector for an electron is \(\vec{r}=(5.0 \mathrm{~m}) \hat{\mathrm{i}}-\) \((3.0 \mathrm{~m}) \hat{\mathrm{j}}+(2.0 \mathrm{~m}) \hat{\mathrm{k}}\). (a) Find the magnitude of \(\vec{r}\). (b) Sketch the vector on a right-handed coordinate system.

A baseball is hit at ground level. The ball reaches its maximum height above ground level \(3.0 \mathrm{~s}\) after being hit. Then \(2.5 \mathrm{~s}\) after reaching its maximum height, the ball barely clears a fence that is \(97.5 \mathrm{~m}\) from where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?

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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.