Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 8

Is there any point on a projectile's trajectory where velocity an acceleration are perpendicular?

Short Answer

Expert verified
Yes, there is one point on a projectile's trajectory where the velocity and the acceleration are perpendicular. This point is the highest point on the path of the projectile.

Step by step solution

01

Understand the Concepts

Firstly, understand that in projectile motion, velocity is a vector quantity that shows the rate of change of displacement, it has a direction and magnitude, and it changes over the trajectory. On the other hand, acceleration due to gravity plays a key role in the projectile's motion and it acts downward throughout the trajectory.
02

Calculate Velocity and Acceleration

Then, consider a standard projectile motion, the velocity vector (\( v \)) changes from the initial velocity vector (\( v_0 \)) to a final velocity vector which is only in downward direction due to gravity (i.e., \( -v_0 \)). The projectile's acceleration (\( g \)) is constant throughout the motion and in opposite direction to the initial velocity vector, since it's due to gravity.
03

Explore the Relationship between Velocity and Acceleration

Next, explore the relationship between velocity and acceleration at any point along the trajectory. If they are perpendicular at any point, then at that point the dot product of \( v \) and \( g \) will be zero, as the dot product of any two perpendicular vectors is zero. But, at any given instant of the projectile's motion, \( v \) and \( g \) always have an angle between them (except at the maximum height where velocity becomes purely horizontal and acceleration is vertical and they are perpendicular to each other).
04

Determine the Perpendicularity

Finally, note that these vectors become perpendicular at the highest point in the trajectory path where the velocity becomes purely horizontal and the acceleration due to gravity is strictly downwards. Hence, these vectors are perpendicular at the highest point of the projectile path and the dot product of velocity and acceleration vectors is zero at this point.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

You wish to row straight across a 63 -m-wide river. You can row at a steady \(1.3 \mathrm{m} / \mathrm{s}\) relative to the water, and the river flows at \(0.57 \mathrm{m} / \mathrm{s} .\) (a) What direction should you head? (b) How long will it take you to cross the river?

A car drives north at \(40 \mathrm{mi} / \mathrm{h}\) for 10 min, then turns east and goes \(5.0 \mathrm{mi}\) at \(60 \mathrm{mi} / \mathrm{h} .\) Finally, it goes southwest at \(30 \mathrm{milh}\) for 6.0 min. Determine the car's (a) displacement and (b) average velocity for this trip.

A carpenter tosses a shingle horizontally off an 8.8 -m-high roof at \(11 \mathrm{im} / \mathrm{s}\). (a) How long does it take the shingle to reach the ground? (b) How far does it move horizontally?

A particle leaves the origin with initial velocity \(\vec{v}_{0}=\) \(11 \hat{\imath}+14 \hat{\jmath} \mathrm{m} / \mathrm{s},\) undergoing constant acceleration \(\vec{a}=-1.2 \hat{\imath}+\) \(0.26 \hat{\jmath} \mathrm{m} / \mathrm{s}^{2} .\) (a) When does the particle cross the \(y\) -axis? (b) What's its \(y\) -coordinate at the time? (c) How fast is it moving, and in what direction?

You're sailboarding at \(6.5 \mathrm{m} / \mathrm{s}\) when a wind gust hits, lasting \(6.3 \mathrm{s}\) accelerating your board at \(0.48 \mathrm{m} / \mathrm{s}^{2}\) at \(35^{\circ}\) to your original direction. Find the magnitude and direction of your displacement during the gust.
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Scientific Method Physics

Read Explanation

Measurements

Read Explanation

Work Energy and Power

Read Explanation

Fluids

Read Explanation
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.

Sign-up for free