This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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From the course by University of Houston System

Preparing for the AP Physics 1 Exam

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University of Houston System

35 ratings

This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

From the lesson

Kinematics

Topics include velocity, acceleration, free fall, projectile problems and graphical analysis. You will watch 3 videos, complete 3 sets of practice problems and take 2 quizzes. Please click on the resources tab for each video to view the corresponding student handout.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

As our experience tells us,

Â not all objects in free-fall travel in just the y direction.

Â Up to this point, we have only analyzed the motion of objects moving vertically.

Â More often than not, an object will move through the air with some displacement in

Â the x and y directions.

Â When an object is launched at an angle, like a cannonball fired at 30 degrees

Â above the x-axis, notice that its velocity is not entirely vertical or horizontal.

Â It is instead a combination of the two.

Â Again, ignoring air friction,

Â this projectile is accelerated downwards towards the center of the Earth due to

Â the force of gravity, which is the only force acting upon it.

Â Since this acceleration vector points in only the y direction.

Â Gravity affects the motion of the cannonball in only the y direction.

Â This allows us to analyze the horizontal, and vertical,

Â directions separately, with different equations along each axes.

Â This leads us to an interesting conclusion.

Â Without air friction, the horizontal velocity for a projectile does not change.

Â So there are no forces acting in the x direction.

Â There can be no acceleration in the x direction.

Â In other words, no change in the speed in the x direction.

Â In the y direction, the cannon ball does experience an acceleration.

Â It begins with its motion after launch with a positive velocity as it

Â travels upward.

Â Due to an acceleration of negative 10 meters per second squared,

Â the canno ball loses speed in the y direction as it

Â approaches its highest point above the ground.

Â At the very top, it has a velocity of zero in the y direction for just an instant.

Â At which point, it turns around and begins to gain speed as it approaches the ground.

Â Separating the motion of your projectile in its x and

Â y directions is a common technique.

Â Notice, however, that we have not yet discussed the total velocity of

Â the cannonball at any point in time as it flies through the air.

Â To do so requires the vector sum, or resultant of the x and y velocities.

Â For instance, the total velocity at this point early in the flight would point up

Â and to the right, but would no longer be oriented at 30 degrees.

Â Because the vertical component of the cannonball's velocity is now smaller

Â than it was at launch.

Â At a later time, such as at the very top of the flight, the total

Â velocity of the projectile is pointed entirely in the horizontal direction.

Â Because its velocity in the y direction is zero at this instant.

Â >> Let's look at an example.

Â A golf ball is launched at 40 degrees with a velocity of

Â ten meters per second above the horizontal.

Â Calculate the total velocity of the golf ball, 0.75 seconds after launch.

Â Next, calculate the maximum range of the projectile when it returns to

Â its starting height.

Â To solve a longer problem like this, it's helpful to be as organized as possible.

Â First I'm going to start with a quick sketch.

Â The golf ball is being launched at an angle of 40 degrees.

Â It's going ten meters per second.

Â The ball will go up, reach some highest point, and

Â come back down, and make a parabola.

Â And, we want to know how fast is it going, and

Â in what direction at 0.75 seconds after the launch, at this point here.

Â So, to get to that we have to start using our vectors.

Â I'm going to break up my launch feed into a horizontal v-naught x and

Â v-naught y components, because we need to have our x and our y components separate.

Â So we can analyze them separately.

Â To do so you can use trig, or if it's to find the y component, I can use sine.

Â Of forty, would equal opposite over hypotenuse, which in this

Â case would be v-naught y, over here, our total launch speed v,

Â and rearranging that, v-naught y will equal v sine of 40.

Â And you'll see us use this shortcut a lot.

Â The hypotenuse times the sine or cosine of the angle we need.

Â So this gives me the y component.

Â The x component, cosine of 40 will equal v-naught x,

Â the adjacent over the hypotenuse.

Â And so my v-naught x will equal v cosine of 40.

Â Those will give me my two horizontal and vertical speeds.

Â I can plug in numbers to get those.

Â V-naught y, and lemme scroll down here for

Â a second so you can see what we're doing.

Â And so my v-naught y will be 10 times the sine of 40,

Â giving me an initial velocity in the y

Â direction of 6.428 meters per second.

Â And that's positive because the object is going to the upward direction at

Â the very beginning of the problem.

Â Same thing with my v-naught x,

Â 10 cosine of 40, giving me

Â a v-naught x of 7.66 meters per second.

Â And, again, positive because the object's going to

Â the right at the beginning of the problem.

Â And notice that our x component is actually slightly bigger than our y

Â component, which makes sense based on the shape of our triangle.

Â And that we're at only 40 degrees, not, let's say, 70 degrees.

Â In which case, in our triangle, you'd have the smaller x component and

Â a larger y component.

Â Okay.

Â We haven't gotten too much further yet

Â because what we really want to know is the final speed.

Â Let me just scroll back up again.

Â At a point here.

Â And there's a couple of things that we need to realize.

Â First the horizontal component of the speed never changes.

Â So this vx.

Â At that location is the same as our initial vx,

Â at the beginning of the problem here.

Â So that hasn't changed, I don't have to do that calculation.

Â We need to find the y component that will either be up or

Â down in this area at that point in time to give me that final speed.

Â So I'm going to come down here and let's look at that.

Â In the y direction, so I'm going to go ahead and

Â label that for myself, vertical, so that I don't confuse which variables I'm using.

Â In the vertical direction,

Â I'm going to use this equation from our kinematic set: vf equals v-nought plus at.

Â And I'm looking for the final velocity that we have.

Â We know in the y direction up here that it started.

Â With 6.428 meters per second.

Â That was a positive value.

Â Then my acceleration's a negative 10 in the y direction due to gravity, and

Â we know that we're looking for this golf ball's speed at 0.75 seconds.

Â And so I get a final velocity in the y

Â direction of negative 1.07 meters per second.

Â That means it's now on its way back down.

Â We've passed the mid-point and we've got a total velocity at that point of,

Â a vector sum in the horizontal direction.

Â That hasn't changed.

Â Adding my vectors tip to tail in the y direction, a negative 1.07.

Â So to find the total velocity here, I need the hypotenuse.

Â And to give a velocity, I also need to find that angle.

Â So this is my right triangle.

Â To find the total velocity v, we're going to take v-final squared

Â will equal v in the x direction squared, plus v in the y direction squared.

Â My distance formula or

Â my Pythagorean theorem, however you'd like to think about that.

Â So v will equal the square root of

Â 7.66 squared plus 1.07 squared.

Â I'm ignoring a negative sign here because it would cancel out and

Â I'm just looking for the magnitude; we'll get direction in a moment.

Â Scrolling down so we have a little bit more room to work.

Â My velocity.

Â In this case just the magnitude of the velocity will be

Â 7.73 meters per second.

Â And as a check, I know my hypotenuse should be

Â bigger than my other two sides of the triangle and that seems to fit.

Â So, so far we seem to be doing pretty well.

Â To find the angle, I need to use trig.

Â So it looks to me like I know my opposite, my adjacent and

Â my hypotenuse, so I could use any trig function.

Â I will go ahead and use tangent.

Â I know my tangent of theta equals my opposite, over my adjacent.

Â And so if I were to take the inverse trig function of this,

Â it'd look something like this.

Â You could call this arctangent, if you aren't familiar with inverse tangent,

Â 7.66 equals theta, so plugging that into my calculator to get the angle,

Â I'll write it out over here to the left.

Â My angle comes out to be 7.95 degrees.

Â Again, I'm not worrying about positive or

Â negative signs here because I can rely on the diagram that we drew above.

Â In my sketch I can see that the projectile is now going below the horizontal.

Â So I want to include that.

Â Below horizontal.

Â It was going to the right, but also going down.

Â The second part of the problem asks us to calculate the range of the projectile.

Â This distance here, x.

Â To do that I'm going to use the horizontal direction.

Â And we had already calculated the initial and final velocities at

Â a specific point and at the very beginning of the problem, but not at the very end.

Â To calculate the total horizontal distance,

Â I'm going to use the horizontal direction.

Â I could use an equation like this,

Â x equals v-naught t plus one-half at squared.

Â But if I'm talking about the horizontal direction,

Â my acceleration in the direction will be zero, so I can ignore that, and

Â use the much more familiar equation, x equals a velocity times time.

Â And since we are splitting these up in the x direction,

Â I know the velocity in that x direction, I just don't know the time.

Â How long were projectiles in the air for the entire trip?

Â That is requiring me to analyze the vertical direction.

Â The trip will be over when it hits the ground.

Â So, to do that,

Â I'm going to pick another kinematic equation, one that will help me solve.

Â To find the time, it could be done with several different equations.

Â I might be able to do x equals v-naught t

Â plus one-half at squared in the y direction.

Â The total displacement at the end of the trip would be zero.

Â I would have to find my initial velocity which we did earlier, 6.4, and

Â the acceleration should be a negative 10 for our motion.

Â I could even choose vf equals v-naught plus a t,

Â there are often many correct choices.

Â Find one that works and, and go that route, whatever you come up with first.

Â I'll go ahead and use the second equation here.

Â There's a lot of symmetry in problems when it comes to projectiles.

Â For instance, like I mentioned earlier,

Â there's symmetry on the time up as well on the way down.

Â It should take the same amount of time to return back to its

Â original height from the very top of its motion here.

Â Also we found this initial v-naught y going upward.

Â That will be the velocity that the projectile has in the y direction on

Â the way back down.

Â So using that symmetry I'm going to come down again and

Â use that piece of information to help me solve.

Â I know at the end of the trip.

Â You will have a negative v sine of theta.

Â It started off

Â with a positive v sine of theta and then we're going to have our acceleration and

Â our time and so I can go ahead and sub in my values for that.

Â A negative, it started with a magnitude of 10.

Â Launched an angle of 40 degrees, 10.

Â again, launched at an angle of 40 degrees

Â minus our 10 acceleration because it's downward, and I'm solving for time.

Â I would move the 10 sine of 40 to the left hand side of

Â the equation by subtracting it.

Â Divide by 10.

Â And another good check here.

Â I'm not going to end up with a negative time.

Â The negative signs will cancel out.

Â You should never end up with a negative time in a problem like this.

Â That would be before the problem started.

Â So my time, according to this calculation, is 1.286 seconds.

Â That's the whole time of flight that I can then come back over here and

Â use in this equation.

Â X will equal, now remember, this is in the horizontal direction.

Â So this velocity will be the velocity that we found earlier that never changed.

Â The 7.66 multiplied by the whole time of flight, 1.286.

Â Will give me a total distance, a total range of this trip of

Â something like 9.851 meters for the golf ball.

Â That's how far it will travel if launched at this speed at this angle.

Â It would go a different distance if the angle or the speed had been changed.

Â Consider now this classic physics example.

Â A gun fires a bullet horizontally in a large open field.

Â At the same instant,

Â a second, similar bullet is dropped from the same height as the one being fired.

Â Which bullet hits the ground first?

Â Let's consider what we've learned so far about projectile motion.

Â We can separate the motion of these bullets into their horizontal and

Â vertical components.

Â The time that these bullets are in the air depends upon what occurs in

Â the y direction.

Â When they hit the ground, the problem is over.

Â Note, too, that the gun did not fire the bullet upward or downward at all.

Â But instead aimed perfectly in the horizontal direction.

Â This means we can completely ignore the horizontal speeds of the bullets,

Â as they do not contribute at all to the time of the fall.

Â Looking more closely at the y direction, then, we see a lot of similarities.

Â Both bullets start with zero initial velocity.

Â Both bullets experience the same negative acceleration to the center of the earth.

Â Both bullets fall from the same height and

Â will therefore reach the ground at the same instant.

Â >> One last example could be helpful in our discussion of accelerated motion in

Â two dimensions.

Â A rock is thrown from the top of a cliff, 100 meters above the sea below.

Â If the rock is launched with an initial velocity of 5 meters per second

Â at an angle of 60 degrees, calculate the time it takes for

Â the rock to hit the water below.

Â Okay, so as we begin a problem like this,

Â I like to first look at what the problem has given me.

Â So let's record our given values.

Â It provides me with the height of the cliff.

Â So my height is 100 meters.

Â It tells me that it's launched with an initial velocity of 5 meters per second at

Â an angle of 60 degrees.

Â So this is being launched at an angle, 60 degrees.

Â I'll label this as my initial velocity.

Â Notice that I haven't put horizontal or

Â vertical, because this velocity is my combined velocity.

Â This is my resultant velocity.

Â Well, that means that I need to solve for my horizontal and

Â vertical components using this velocity.

Â So, let's look at that real quick.

Â So, if I have 5 meters per second here, I'm going to look at this as if

Â it's a triangle right here, here is the horizontal, here is the vertical.

Â I want to solve for the horizontal component.

Â This would be my v of x.

Â And then I want to solve for the vertical component.

Â This would be my v of y.

Â Well you thi, you think back to your trig.

Â In this case, I have the hypotenuse.

Â In order to get my v of y,

Â my vertical velocity, I need to solve for the opposite side.

Â In order to get my v of x, the horizontal, I need to solve for the adjacent side.

Â So let's solve for the v of y first.

Â When I'm solving for the v of y, knowing that I have the hypotenuse and

Â I need the opposite, I'm going to use sine.

Â So v of y is going to equal v sine of theta, while v of x will equal v,

Â and that's the initial velocity, cosine of theta.

Â Let's go ahead and plug values in here.

Â So, my horizontal is going to be 5 cosine of 60 degrees.

Â My initial vertical is going to be 5 sine of 60 degrees.

Â Notice how I didn't put an initial or a final next to the horizontal.

Â I did this because the horizontal velocity is going to stay constant.

Â So there's no point in me putting initial horizontal velocity versus final.

Â There is a point in doing that for the vertical.

Â Because the vertical component is going to change due to acceleration,

Â due to gravity acting upon it.

Â So, in this case my horizontal, when I plug this in,

Â comes out to be 2.5 meters per second.

Â Vertical component comes out to be 4.33 meters per second.

Â Looking back at the vector, is it positive, or negative vertical component?

Â Well, this vertical component is pointing upwards, so

Â it's going to be a positive velocity.

Â If you're not getting the values that I have over here,

Â you may want to check the mode in your calculator.

Â I am currently using degree mode.

Â So let's go ahead and continue.

Â I want to solve for the total time, the total time of this object's flight.

Â Well, I'm going to show you two different ways to do this.

Â Here's the first way.

Â I have the height.

Â The height is 100 meters.

Â Say I want to use my kinematic equation.

Â Y equals one-half at squared plus v initial

Â in the vertical direction times t.

Â I had to use one of my kinematics equations that involves time,

Â and since I can figure out what y is, this would be a good equation to start with.

Â Well, let's look at what y is.

Â A common misconception might be that y,

Â the displacement in the vertical direction, is 100 meters.

Â Well, but that's the height of the object.

Â When this object, say that this point right here is my y equals 0.

Â From y equals 0, say this object is returning back to the ground,

Â this displacement, this change in position is actually negative.

Â So my displacement, my vertical displacement,

Â is actually going to be negative 100.

Â Let's go ahead and plug that in.

Â So I have negative 100 meters for my vertical displacement,

Â one-half a, a being the acceleration due to gravity, negative 10,

Â multiplied by time squared, plus my initial vertical component

Â which was the positive 4.33, multiplied by time.

Â Just a quick reminder, I would not be able to use the horizontal component of

Â velocity over here because the horizontal component is not

Â being acted upon by the force of gravity.

Â It does not have acceleration due to gravity acting on it.

Â So we wouldn't be able to use an equation that involves acceleration in it.

Â All right.

Â Let's go ahead and continue from here.

Â Now in order for me to solve for this, I'm going to add t squared and t.

Â These are my two.

Â My one unknown over here that's being repeated.

Â Some of you may already know where I'm going to go with this.

Â I'm going to solve for this using the quadratic.

Â So, I need to place this in the standard form.

Â That is ax squared plus bx plus

Â c equals 0.

Â So then I can go ahead and rewrite this my ax squared, so

Â that would be this one-half times negative 10, giving me negative 5 t squared.

Â The b that I have is 4.33.

Â This will end up becoming, negative 4.33 due to the rearrangement.

Â The c is negative 100, so I've moved all these variables

Â over to one side, setting them all equal to 0.

Â Now I'm ready to start using my

Â quadratic equation, which if you recall,

Â is the negative b plus or minus root

Â b squared plus 4ac divided by 2a.

Â So then I can go ahead.

Â And plug values in over here.

Â Negative b, well, my b is negative 4.33 so

Â I can just write that out as 4.33, plus or minus,

Â root 4.33 squared plus 4 times a,

Â is negative 5, times c, which is negative 100.

Â All of this is being divided, right here by 2 times a.

Â Again my a is negative, 5.

Â So when I solve for this I'm going to get two values.

Â The two values that I end up with,

Â I get t equals 4.93, and

Â I get t equals negative 4.06.

Â So this second value is not the value that I'm going to use.

Â The second value's actually the solution corresponding to

Â a time before this rock was thrown, so it's not going to apply here.

Â I'm going to use time equals 4.93, so my final answer, for

Â this question is 4.93 seconds.

Â And that's what I got using the quadratic.

Â There is another way to do it.

Â It's a little bit longer and

Â it involves some of our kinematics equations coming together.

Â So I'm going to show you how to do that as well.

Â It's better to know how to solve a problem more ways than less, so let's go ahead and

Â look at this one again.

Â Once again I've got the same stuff provided to me, I have the initial

Â velocity, 5 meters per second, it is at an angle of 60 degrees from the horizontal.

Â Height of 100 meters, displacement in the vertical direction being negative 100.

Â So, let me go ahead and see how I'm going to solve for this one.

Â The way that I'm going to do this,

Â I'm going to break this apart into a couple of pieces here.

Â So, starting from right here.

Â I can solve for the time just of this path over here that I've outlined.

Â If I can solve for this time first, then I can continue to solve for

Â the remaining problem and add the two times up together.

Â So let me go ahead and do that.

Â I know that my initial velocity over here is 5 meters per second.

Â It's got a vertical component.

Â My initial vertical velocity was 4.33,

Â oops, 4.33 meters per second.

Â Let's re-write that.

Â Perfect. Okay, now let's go ahead and solve for

Â the time that it takes for it to reach this point.

Â Well, remember, my symmetry still applies here.

Â This final velocity vector right over here at this point is also going to be

Â the 5 meters per second, as it's approaching.

Â So the symmetry applies and my final velocity over here is also

Â the 5 meters per second, which means that I, again, have the same

Â vertical component pointing downwards, so this time, at this point right here,

Â my final vertical is going to be negative 4.33 meters per second.

Â The magnitudes are the same, the direction is different because one vector is

Â pointed upwards, the other vector is pointed downwards.

Â From here I can solve for the time.

Â a good equation to use over here would be vf equals v-initial plus at.

Â I'm looking at this in the vertical, so

Â this is final velocity vertical equals initial velocity vertical plus g,

Â acceleration due to gravity, multiplied by time.

Â And scroll down and plug some values in for this one.

Â So I have final velocity of negative 5 equals initial velocity positive

Â 5 plus the negative 10 for acceleration, and then I can solve for time.

Â I end up getting 0.866 seconds for time.

Â Okay, so now I know how long it took for the object to just reach that one point.

Â Now what I need to do is I need to go back over here and I need to solve for

Â the time that it takes this object going from here on down.

Â So, in order to solve for that, what I need to do now is to

Â figure out what velocity this object has when it reaches the water.

Â So, I want to figure out what this final velocity is over here.

Â Well, I know what my initial velocity was,

Â again, starting from the point where we left off at that 0.866 seconds.

Â So let's go ahead and use another one of our kinematics equations.

Â This time I'm going to use the equation

Â vf squared in the vertical direction equals v initial squared in

Â the vertical direction plus 2, the vertical acceleration.

Â That's g, due to gravity.

Â Multiplied by y, the vertical displacement.

Â The vertical displacement, remember, is the negative 100 meters.

Â So I can go ahead and plug in my values.

Â Start solving for this.

Â My initial vertical velocity.

Â Was the negative 4.33.

Â Square that, plus 2 times negative 10,

Â multiplied by negative 100.

Â When I solve for

Â this final vertical component I end up with 44.93.

Â Now I can go back and use another equation that I've already used which is.

Â I'm going to come back over here.

Â And I will use vfy equals v-initial plus at.

Â I'm going to solve for time, here.

Â And once I get this time, I can go ahead and

Â combine the two times together to get my total answer.

Â So I will now have the final vertical, which is the 44.93.

Â It is negative and I'm going to go ahead and add that negative over here too.

Â I also know it's negative because of the fact that it is pointing downwards.

Â It's coming to the ground.

Â Initial is the negative 4.3, plus negative 10 multiplied by time.

Â And when I solve for this time over here and

Â I add this time value with

Â the 0.866 seconds, my time total

Â comes out to be 4.926 seconds.

Â That's the same total time that I had earlier.

Â Using the quadratic equation.

Â So again, this method is longer but if you absolutely do

Â not want to use the quadratic then you can still use your kinematics equations.

Â And this also gave us some practice utilizing some of our equations such as

Â the vf squared equals v-initial squared plus 2ax that

Â we hadn't really used much of before.

Â >> One last note before we end this module on kinematics.

Â You will sometimes encounter problems in which an object is released or

Â launched out of a moving vehicle.

Â For example, a package dropped out of a flying airplane.

Â Such a set-up gives you information regarding the initial speed and

Â direction of the object.

Â For instance, suppose that a plane is flying at 270 meters

Â per second at 10 degrees above the horizontal.

Â A package at rest relative to the plane is then also moving at 270 meters

Â per second at 10 degrees above the horizontal.

Â If the package is then released from the cargo hold so

Â that it is not given any additional velocity,

Â the package starts its motion with this initial speed and direction.

Â There is no difference between a package released in this manner

Â versus launching it at this velocity using other means.

Â The subsequent problem solving techniques would be the same.

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