It’s a long fly ball to center field… in this episode, we take a look at projectiles.
It’s a long fly ball to center field… in this episode, we take a look at projectiles. We identify some of the assumptions we can make about projectiles (1:13), and then we take a look at projectiles launched horizontally, (3:37) and we solve for the horizontal range (5:28). The episode then takes a look at a projectile launched at an angle and figures out where a basketball player should shoot from (7:22). We bring it on home by pointing out some nuances and fun facts about projectiles (9:54).
The Question of the Day asks: (11:30)
A projectile launched from level ground at an angle of 20 degrees will have the same range as a projectile with the same launch velocity at ___ degrees.
a) 80 b) 70 c) 60 d) 50
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Hi and welcome to the APsolute Recap: Physics 1 Edition. Today’s episode will focus on projectile motion problems.
Lets Zoom out:
Unit 1 – Kinematics
Topic 1.1
Big Idea – Force Interactions
A homerun at a baseball game, a long pass in soccer, a drive down the fairway in golf, what do they all have in common? They all move in two dimensions. More commonly they are referred to as projectiles, but what does that mean and how can we predict their motion with what we already know about one dimensional motion? The good news… no new formulas! The… ummm… not as good news, we will be keeping track of two motions at once.
Lets Zoom in:
A projectile is an object experiencing free-fall, but with some initial velocity in the horizontal, x direction. Remember that to be in free-fall means that only gravity is causing it to accelerate. Projectiles always follow the same type of path as they move in two dimensions, they always have a parabolic path, meaning that if we trace their motion through the air it would be a parabola.
A marksman aiming down range with the barrel of his gun held horizontally fires at the target. At the moment that the bullet leaves the marksman’s gun, another object is dropped from the same height above the ground. Which hits the ground first? The answer is that they both hit at the same time since they both had an initial velocity in the y-direction equal to 0 m/s.
A ball is placed at rest on a level table. Does the ball ever randomly begin speeding up in either the forward or reverse direction? No, of course not. That is because there is not an acceleration due to gravity that acts in the x-direction, only the 9.8 m/s2 acceleration which acts in the negative y-direction.
While at a football game, you are seated behind the field goal post and watch the opening kickoff. You see the ball leave the ground and travel skyward. It rises while slowing, it stops at its peak, and then it speeds up in the negative direction. Seems a whole lot like a ball thrown up into the air and then caught at the same height. You tell your friend over the phone about how it looks just like a one dimensional free-fall problem you remember hearing about while listening to The APsolute recap Physics 1 edition the other day. Your friend is at home watching the game and says, “That’s ridiculous, because I am watching the game at home and they just showed an overhead view of the kick off and the ball appears to travel downfield at a constant velocity as it travels past the 30, 40, and 50 yd lines all the way down the field.” Which one of you is correct? You - watching from behind the field goal post, or your friend - who is watching an overhead view on TV? So, as it turns out you and your friend are both correct since projectiles only accelerate in the y-direction.
Let’s try to solve an example problem in which an object is launched horizontally. It’s summertime, and you are at the local swimming pool when you run off the diving board horizontally with a velocity of 7 m/s and eventually splash into the water some 10 m below you. How far from where you leapt horizontally off the diving board should your friend have an inner tube waiting so that you can jump right through it like you are a performer in a 3 ringed circus?
The best way to organize all of this information coming at you in two dimensions is to create a table. This will help you keep track of the relevant information in both the x and the y directions. The relevant information in each of those directions is the acceleration, time, initial and final velocities and the displacement. I envision this table having two columns - one for the x-direction and one for the y-direction. Acceleration in the x will be 0 m/s2 and while on earth, the acceleration in the y direction will be -9.8 m/s2. The time we are currently unsure of so we will leave both columns blank, but it is important to point out that projectiles only fall once and they do so in both directions at the same time. The initial velocity in the x direction will be 7 m/s, but in the y-direction it is not so straightforward. We know that you left the diving board horizontally, which indicates that the initial y-velocity should be 0 m/s. We also know that in the x-direction your final velocity is also 7 m/s since you won’t accelerate in the x-direction. Finally, the displacement in the x-direction is what we ultimately are looking for, and in the y-direction we will place a -10 m since you moved in the downward direction from your starting location.
Now we have to figure out the best equation to use for what we know. It appears the 2nd kinematic equation from the Physics 1 equation sheet is the right choice, but we should be reminded that we can only place values into that equation that come from the same side of our table we have created. In this case, we will work with the y-direction first and we can set the displacement equal to ½ times the acceleration multiplied by the time squared. We should have -10 m = -4.9 m/s2 times “t” squared. We find that you are airborne for 1.4 seconds. Armed with this information we can turn our attention to the x-direction side of our table and use the 2nd kinematic equation again. This time we have displacement equals the initial velocity times the time. In your case, the displacement equals 7 m/s times 1.4 s or 9.8 m. Pretty straight-forward right?
Now let’s take a peek at a projectile launched at an angle. A basketball player shoots the ball with a speed of 10 m/s at an angle of 60 degrees above horizontal into a net located a meter above where it is shot from. Maybe we want to find out where the player needs to stand in order to make the basket.
Again, we should create an “a, t, v” table and fill in what is known. The standards are still in play with respect to the acceleration in each direction, so again acceleration in the x direction is 0 m/s2, and the y-direction acceleration is -9.8 m/s2 on Earth. Time is unknown and hopefully, you can see that being shot at an angle gives the ball velocity in both the x and y directions, so the 10 m/s is the resultant of the x and y components of velocity. With a little trig you can solve for each and find that the initial x velocity is 5 m/s in the forward direction and the y velocity is 8.7 m/s in the up direction. Additionally, you know the y displacement is +1m since the ball ends its path 1 m above where it started.
One route to solving is to look at the y-column of the table you created and to use the 3rd kinematic equation to find the final y direction velocity, but you’d have to recognize that the answer can be either positive or negative since the final velocity will get squared. But, if we know a little bit about basketball we know the ball has to enter the top of the net, and the ball must have a negative velocity of -7.5 m/s to do that. We can then use the 1st kinematic equation to solve for the ball’s hangtime which ends up being 1.65 seconds.
OR… we could have used the 2nd kinematic equation and the quadratic formula to solve for the time directly, and then choose the later time when the ball has had time to go up 1 m, reach its peak, and then come back down to a height of 1 m above the ball’s starting point. Either way, we get the time of 1.65 seconds.
Lastly, you know the x velocity and the time, so the 2nd kinematic equation can be used to take the initial x velocity of 5 m/s multiplied by the hangtime of 1.65 seconds to find that the basketball player should take the shot when he is a horizontal distance from the basket.
A few last things! The time to the peak position cannot be assumed to always be half the hangtime, BUT… when a projectile is launched and it lands at the same location in the y-direction then the displacement in the y-direction is 0 m and the time to the peak will be half the total hang time. Again, that’s a special case! So, be careful using it. Another fun fact when the y-displacement is 0 m, then projectiles launched at angles that are complementary (add to 90 degrees) will have the same range! Shoot a ball at 30 degrees or 60 degrees above horizontal, and they will land at the same place on level ground. Lastly, and this is a big one… throughout the entire journey that a projectile makes through the air, the acceleration is downward and equal to -9.8 m/s2. Gravity doesn’t suddenly turn off at the peak!
To recap……
Projectiles in free-fall are only accelerated by gravity. Acceleration in the x direction is 0m/s2 if in free-fall, and -9.8 m/s2 in the y direction. When launched at an angle, the x and y components of velocity can be found using some simple right angle trig. The y velocity is 0 m/s at the peak, and with the same 3 kinematic equations we used in 1-D motion, you can solve any projectile problem.
Coming up next on the APsolute RecAP Physics 1 Edition, we will focus on dynamics and the forces that exist in our world.
Today’s Question of the Day focuses on projectile motion.
Question: A projectile launched from level ground at an angle of 20 degrees will have the same range as a projectile with the same launch velocity at what angle?
a) 80 b) 70 c) 60 d) 50