We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

Create An Account

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #35 : Understanding Motion In Two Dimensions

Laurence throws a 0.5kg rock off the edge of a 12m tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the total horizontal distance that the rock travels?

Possible Answers:

86.81m

21.45m

101.2m

52.71m

160.2m

Correct answer:

86.81m

Explanation:

To find the horizontal distance, we will need to use the horizontal velocity and the total time that the rock is in the air. We can easily find the horizontal velocity by using the cosine function and the initial velocity and angle.

$v_{i}* \cos(\theta)=v_{x}$

$31\frac{m}{s}* \cos(20^{\circ})=v_{x}$

$31\frac{m}{s}*(0.94)=v_{x}$

$29.13\frac{m}{s}=v_{x}$

There is no horizontal acceleration, so the velocity will be constant. This means our equation for distance will be simply:

$\Delta x =v_xt$ .

We know v_x , but now we need to find the time the rock is in the air.

Use the sine function with the initial velocity and angle to find the vertical velocity.

$v_{i}*\sin(\theta)=v_{iy}$

$31\frac{m}{s}*\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}* (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

We need to break the rock's path into two parts. The first part is the time that the rock is rising to its maximum height, and the second part is the time that it is falling from the highest point to the ground.

For the first part, we can assume that the final vertical velocity is zero, since this will be the top of the parabola. Using the initial velocity, final velocity, and gravity, we can solve for the time to travel this portion of the path.

v_f=v_i+at_1

$0\frac{m}{s}=10.6\frac{m}{s}+(-9.8\frac{m}{s^2})t_1$

$-10.6\frac{m}{s}=-9.8\frac{m}{s^2}\cdot t_1$

$\frac{-10.6\frac{m}{s}}{-9.8\frac{m}{s^2}}=t_1$

1.08s=t_1 .

This is the time that the rock is traveling upward. Now we need to focus on the time that the rock travels downward. We don't know the final velocity at the end of the parabola, so we can't use the equation from the first part. If we can find the total height of the parabola, we can use a different equation to solve for time.

$\Delta y =v_it+\frac{1}{2}at^2$

Remember that the vertical velocity at the highest point of a parabola is zero. Now that we know the initial vertical velocity, we can plug this into an equation to solve for the distance that the rock travels from its maximum height to the original height.

$v_f^2=v_i^2+2a\Delta y$

$(0\frac{m}{s})^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$

$(0\frac{m}{s})=(112.36\frac{m}{s})+(-19.6\frac{m}{s^2})\Delta y$

$-(112.36\frac{m}{s})=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-112.36\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$5.73m=\Delta y$

Remember, $\Delta y$ only tells us the CHANGE in height. Since the rock started at the top of a 12m tall building, if it rose an extra 5.73m , then at its highest point it is 17.73m above the ground.

This means that our $\Delta y$ for the second equation will be -17.73m (the change in height is negative because it travels downward.) Use this total distance and the velocity at the top of the peak (zero) to solve for the time that the rock travels down.

$\Delta y =v_it+\frac{1}{2}at^2$

$-17.73m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-17.73m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-17.73m=(-4.9\frac{m}{s^2})t^2$

$\frac{-17.73m}{-4.9\frac{m}{s^2}}=t^2$

3.62s^2=t^2

$\sqrt{3.62s^2}=\sqrt{t^2}$

1.90s=t_2

Finally, add the two times together to find the total time in flight.

$t_{total}=t_1+t_2$

$t_{total}=1.08s+1.90s$

$t_{total}=2.98s$

Now that we have finally found our time, we can plug that back into the equation from the beginning of the problem:

$\Delta x =v_xt$

Use the time and horizontal velocity to find the total horizontal distance.

$\Delta x =29.13\frac{m}{s}*2.98s$

$\Delta x=86.81m$

Report an Error

Example Question #34 : Understanding Motion In Two Dimensions

Laurence throws a 0.5kg rock off the edge of a 12m tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the total final velocity of the rock right as it hits the ground?

Possible Answers:

$13.23\frac{m}{s}$

$3.46\frac{m}{s}$

$68.21\frac{m}{s}$

$19.69\frac{m}{s}$

$34.58\frac{m}{s}$

Correct answer:

$34.58\frac{m}{s}$

Explanation:

First, break our given velocity into its horizontal and vertical components.

We can find the horizontal velocity using the cosine function, initial velocity, and angle.

$v_{i}*\cos(\theta)=v_{x}$

$31\frac{m}{s}*\cos(20^{\circ})=v_{x}$

$31\frac{m}{s}*(0.94)=v_{x}$

$29.13\frac{m}{s}=v_{x}$

We can find the vertical velocity using the sine function, initial velocity, and angle.

$v_{i}* \sin(\theta)=v_{iy}$

$31\frac{m}{s}*\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}* (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

The horizontal velocity is not going to change, since there is no horizontal acceleration. The vertical velocity, however, will be affected by gravity.

We know that the rock is going to travel a net displacement of -12m , as that's the distance between the rock's initial position (the top of the building) and the ground. The displacement must be negative because the rock is traveling downward. Using the appropriate motion equation, we can use the rock's initial velocity, displacement, and acceleration to find the final velocity.

$v_f^2=v_i^2+2a\Delta y$

$v_f^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-12m)$

$v_f^2=(112.36\frac{m^2}{s^2})+235.2\frac{m^2}{s^2}$

$v_f^2=347.56\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{347.56\frac{m^2}{s^2}}$

$v_f=18.64\frac{m}{s}$

Keep in mind that square roots can be positive or negative. Since the rock is traveling in the downward direction, we need to take the negative root.

$v_f=-18.64\frac{m}{s}$ .

We now have both the final horizontal and final vertical velocities. Use the Pythagorean Theorem to find the total velocity from the directional velocities.

$(-18.64\frac{m}{s})^2+(29.13\frac{m}{s})^2=(v_{R})^2$

$347.56\frac{m^2}{s^2}+848.56\frac{m^2}{s^2}=(v_{R})^2$

$1196.2\frac{m^2}{s^2}=(v_{R})^2$

$\sqrt{1196.2\frac{m^2}{s^2}}=\sqrt{(v_{R})^2}$

$34.58\frac{m}{s}=v_{R}$

Report an Error

Example Question #35 : Understanding Motion In Two Dimensions

Laurence throws a 0.5kg rock off the edge of a 12m tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the angle that the rock will make with the ground right before it hits the ground?

Possible Answers:

$13.62^{\circ}$

$61.08^{\circ}$

$0.32^{\circ}$

$32.62^{\circ}$

$12.45^{\circ}$

Correct answer:

$32.62^{\circ}$

Explanation:

To find the angle, we will need to find the final horizontal and vertical velocities. First, break our given velocity into its horizontal and vertical components.

We can find the horizontal velocity using the cosine function, initial velocity, and angle.

$v_{i}*\cos(\theta)=v_{x}$

$31\frac{m}{s}*\cos(20^{\circ})=v_{x}$

$31\frac{m}{s}*(0.94)=v_{x}$

$29.13\frac{m}{s}=v_{x}$

We can find the vertical velocity using the sine function, initial velocity, and angle.

$v_{i}* \sin(\theta)=v_{iy}$

$31\frac{m}{s}*\sin(20^{\circ})=v_{iy}$

$31\frac{m}{s}* (0.34)=v_{iy}$

$10.6\frac{m}{s}=v_{iy}$

The horizontal velocity is not going to change, since there is no horizontal acceleration. The vertical velocity, however, will be affected by gravity.

$v_f^2=v_i^2+2a\Delta y$

$v_f^2=(10.6\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-12m)$

$v_f^2=(112.36\frac{m^2}{s^2})+235.2\frac{m^2}{s^2}$

$v_f^2=347.56\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{347.56\frac{m^2}{s^2}}$

$v_f=18.64\frac{m}{s}$

Now that we have the final directional velocities, we can use the tangent function to find the angle with which the rock impacts the ground.

$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$

$\tan({\theta})=\frac{v_y}{v_x}$ .

$\theta=\tan^{-1}(\frac{v_y}{v_x})$

$\theta=\tan^{-1}(\frac{18.64}{29.13})$

$\theta=\tan^{-1}(0.64)$

$\theta=32.62^{\circ}$

Report an Error

Example Question #71 : Motion And Mechanics

A potted plant falls off a window sill from 3 meters above the ground. How long does it take before the plant hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

0.782s

0.818 s

1.2 s

0.656 s

0.723 s

Correct answer:

0.782s

Explanation:

$\Delta y = V_i(t)+\frac{1}{2}at^2$

Using the equation above and the given values, we can solve for the time. First we need to find the change in distance.

$\Delta y=y_f-y_i=0m-3m=-3m$

The plant travels $\small -3$ meters in the downward direction. Now we can plug in the given values from the question to solve for time.

$-3m= (0\frac{m}{s})+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-3m = (-4.9\frac{m}{s^2})t^2$

$\frac {-3m}{-4.9\frac{m}{s^2}}=t^2$

0.612s^2=t^2

$\sqrt{0.612s^2}=\sqrt{t^2}$

0.782s=t

Report an Error

Example Question #72 : Motion And Mechanics

Walter throws a disc from 1.5 meters above the ground with purely horizontal motion. How long will it be before the disc hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

0.553 s

1.03 s

0.002 s

0.343 s

0.053 s

Correct answer:

0.553 s

Explanation:

The horizontal motion will not affect the time the disc is in the air. Time will be determined by the rate at which the disc falls: the acceleration due to gravity.

We can solve for the time using the equation below and the values given in the question.

$\Delta y=V_i(t)+\frac{1}{2}at^2$

$(0m-1.5m)=(0\frac{m}{s})(t)+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$(-1.5m)=\frac{1}{2}(-9.8\frac{m}{s^2})t^2\boldsymbol{}$

$(-1.5m)=(-4.9\frac{m}{s^2})t^2$

$\frac{-1.5m}{-4.9\frac{m}{s^2}}=t^2$

0.306s^2=t^2

$\sqrt{0.306s^2}=\sqrt{t^2}$

0.553s=t

Report an Error

Example Question #73 : Motion And Mechanics

Leslie rolls a ball out of a window from 10 meters above the ground, such that the initial y-velocity is zero. What will the ball's final velocity be right before it hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

$12\frac{m}{s}$

$14\frac{m}{s}$

$28\frac{m}{s}$

$9\frac{m}{s}$

$13.2\frac{m}{s}$

Correct answer:

$14\frac{m}{s}$

Explanation:

We are given the initial velocity, acceleration, and the initial and final distances.

$\Delta y=y_f-y_i=0m-10m=-10m$

Using the equation below and the given values, we can solve for the final velocity.

$V_f^2=V_i^2+2a\Delta y$

$V_f^2=(0\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)$

$V_f^2=196\frac{m^2}{s^2}$

$\sqrt{V_f^2}=\sqrt{196\frac{m^2}{s^2}}$

$V_f=14\frac{m}{s}$

Report an Error

Example Question #1 : Using Motion Equations

A 2.9kg crate slides across a floor for before coming to rest 10m from its original position.

What is the initial velocity of the crate?

Possible Answers:

$8\frac{m}{s}$

$1\frac{m}{s}$

$2\frac{m}{s}$

$50\frac{m}{s}$

$4\frac{m}{s}$

Correct answer:

$4\frac{m}{s}$

Explanation:

The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

$\Delta x=\frac{(v_f+v_i)}{2}\Delta t$

Plug in our given values and solve.

$10m=\frac{(0\frac{m}{s}+v_i)}{2}5s$

Divide both sides by $\small 5$ .

$\frac{10m}{5s}=\frac{v_i}{2}$

$2\frac{m}{s}=\frac{v_i}{2}$

Multiply both sides by $\small 2$ .

$4\frac{m}{s}=v_i$

Report an Error

Example Question #2 : Using Motion Equations

A 2.9kg crate slides across a floor for before coming to rest 10m from its original position.

What is the net acceleration on the crate?

Possible Answers:

$-4\frac{m}{s}$

$-0.8\frac{m}{s^2}$

$-2.32\frac{m}{s^2}$

$-1\frac{m}{s^2}$

$-10\frac{m}{s}$

Correct answer:

$-0.8\frac{m}{s^2}$

Explanation:

All of the equations regarding acceleration require an initial velocity. We will need to solve for the initial velocity using the given variables. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

$\Delta x=\frac{(v_f+v_i)}{2}\Delta t$

Plug in our given values and solve.

$10m=\frac{(0\frac{m}{s}+v_i)}{2}5s$

$\frac{10m}{5s}=\frac{v_i}{2}$

$2\frac{m}{s}=\frac{v_i}{2}$

$4\frac{m}{s}=v_i$

We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.

$\Delta x=v_it+\frac{1}{2}at^2$

Plug in the given values to solve for acceleration.

$10m=(4\frac{m}{s}) (5s)+\frac{1}{2}a(5s)^2$

$10m=(20m)+\frac{1}{2}a(25s^2)$

$-10m=\frac{1}{2}a(25s^2)$

-20m=a(25s^2)

$\frac{-20m}{25s^2}=a$

$-0.8\frac{m}{s^2}=a$

Report an Error

Example Question #4 : Using Motion Equations

A ball starts rolling at $1.5\frac{m}{s}$ . It accelerates at a constant rate of $2\frac{m}{s^2}$ for 5.23s . What is the final velocity?

Possible Answers:

$10.46\frac{m}{s}$

$1.5\frac{m}{s}$

$4.12\frac{m}{s}$

$11.96\frac{m}{s}$

$8.22\frac{m}{s}$

Correct answer:

$11.96\frac{m}{s}$

Explanation:

To solve for the final velocity, remember that the relationship between velocity, acceleration, and time is v_f=v_i+at .

Using the given values for the initial velocity, acceleration, and time, we can solve for the final velocity.

$v_f=1.5\frac{m}{s}+(2\frac{m}{s^2}*5.23s)$

$v_f=1.5\frac{m}{s}+10.46\frac{m}{s}$

$v_f=11.96\frac{m}{s}$

Report an Error

Example Question #5 : Using Motion Equations

A book, starting at rest, falls off of a table. What is its velocity after 2.1s in motion?

$g=-9.8\frac{m}{s^2}$

Possible Answers:

$-10.26\frac{m}{s}$

$-20.58\frac{m}{s}$

$-11.9\frac{m}{s}$

$-4.67\frac{m}{s}$

$-7.7\frac{m}{s}$

Correct answer:

$-20.58\frac{m}{s}$

Explanation:

We can solve this question using the equation for acceleration in terms of velocity:

$a=\frac{v_2-v_1}{t}$

We know our initial velocity (zero, since we start from rest), time, and the acceleration of gravity. Use these values to isolate the variable for the final velocity.

$-9.8\frac{m}{s^2}=\frac{v_2-0\frac{m}{s}}{2.1s}$

$(-9.8\frac{m}{s^2})(2.1s)=v_2$

$-20.58\frac{m}{s}=v_2$

Note that the final velocity is negative, since the book is traveling in the downward direction.

Report an Error

View Tutors

Armon
Certified Tutor

Loyola University-Chicago, Bachelor of Science, Biology, General. Columbia College-Chicago, Bachelor of Science, Graphic Desi...

View Tutors

Devan Renea
Certified Tutor

Morgan State University, Associate in Arts, Theater Arts. Full Sail University, Bachelor in Arts, Film Studies.

View Tutors

Annette
Certified Tutor

University of Maryland-University College, Bachelor of Science, Computer and Information Sciences, General. University of Mar...

All High School Physics Resources

6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

English Tutors in Dallas Fort Worth, GRE Tutors in Atlanta, Calculus Tutors in Chicago, MCAT Tutors in New York City, Algebra Tutors in Denver, Biology Tutors in San Francisco-Bay Area, SAT Tutors in Miami, ISEE Tutors in Seattle, French Tutors in Philadelphia, MCAT Tutors in Phoenix

Popular Courses & Classes

SAT Courses & Classes in Chicago, ISEE Courses & Classes in Philadelphia, LSAT Courses & Classes in Houston, GRE Courses & Classes in Houston, ACT Courses & Classes in Seattle, Spanish Courses & Classes in Washington DC, LSAT Courses & Classes in Atlanta, Spanish Courses & Classes in New York City, LSAT Courses & Classes in Washington DC, SSAT Courses & Classes in New York City

Popular Test Prep

ISEE Test Prep in Phoenix, GRE Test Prep in Atlanta, GMAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Los Angeles, SSAT Test Prep in New York City, SAT Test Prep in Denver, GMAT Test Prep in Atlanta, GRE Test Prep in Houston, GRE Test Prep in Denver, GMAT Test Prep in Denver

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #35 : Understanding Motion In Two Dimensions

Example Question #34 : Understanding Motion In Two Dimensions

Example Question #35 : Understanding Motion In Two Dimensions

Example Question #71 : Motion And Mechanics

Example Question #72 : Motion And Mechanics

Example Question #73 : Motion And Mechanics

Example Question #1 : Using Motion Equations

Example Question #2 : Using Motion Equations

Example Question #4 : Using Motion Equations

Example Question #5 : Using Motion Equations

All High School Physics Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: