Question

Projectile motion: Given this data, find the initial velocities vox and voy, predicted horizontal distance and time of flight?

Answer 1

Also what kinematics equation to use for horizontal distance?

Answer 2

Do you know the formulas? Did you try just putting numbers in them? This should be fairly straightforward albeit a bit repetitive.
$$ \vec v = v_x\hat i + v_y\hat j\\
v_{x0} = v_0 \cos\theta\\
v_{y0} = v_0 \sin\theta\\$$I said x0 and y0 because it depends sort of an what exactly are you taking in account. But in principle the vx is constant throughout if you ignore the friction from air. Vy changes because there's the force of gravity acting on the body to bring it back down. Back to classical mechanics and the general formulas for constant accelerated movement
$$v_y = v_{y0} - gt\\
y = v_{y0}t - \frac{1}{2}gt^2$$

Answer 3

yes i know the kinematic equations except i'm not sure how to put it together to solve for each one that i asked for above. I got this from lab where we launched a ball in three different angles (15, 30, 45).

So if i solve for t by using one of those equations, it will give me the time of flight? We're still not worried about air resistance or anything like that so i assume that vx does stay constant. So does that make vx 0?

Answer 4

@ljetibo

Answer 5

That makes vx0 == vx all the time. So if you i.e. knew the initial velocity in the x direction you could've calculated the time of flight. But of course this isn't given. In fact not a lot was in this task. You're going to give a bit more clues about what is going on to be able to solve for time as is. A diagram of the experimental apparatus would be great help.

But as it stands it doesn't particularly matter. We can't solve for time t but we don't have to either, we need to solve for v, or maximal height, or time to maximal height or basically any additional info. In general you can find the range of the projectile to be only related to the initial velocity and angle. Since angle is given, we know the range, we can calc the velocity. First the final formula:
$$R = \frac {v_0\cos \theta} {g} \left ( v_0\sin \theta + \sqrt{v_0^2 \sin^2 \theta + 2 g y_0} \right)$$
You can try with this formula, I can show you how I got it. Reverse it and try getting v0 out of it. It's not going to be really easy though.

Answer 6

Turned out to be a bit easier to flip the formula than expected, getting to it wasn't hard but takes a bit of writing. Once flipped:
$$v_0^2 = \frac{R^2}{\cos\theta^2 \left(2*R\tan\theta + 2y_0\right)}$$
once you put in the stuff you know
$$\cos(15) \equiv \cos \frac{\pi}{12} = \frac{\sqrt6 + \sqrt2}{4}\\
\rightarrow \cos^2 \frac{\pi}{12} = \frac{2 + \sqrt3}{4}\\
\tan(15) \equiv \tan \frac{\pi}{12} = 2 - \sqrt3$$
so:
$$v_0^2 = \frac{4R^2}{ (2+ \sqrt3)\left(2*(2 - \sqrt3)R + 2y_0\right)}$$
but as we know:
$$(2+ \sqrt3)(2 - \sqrt3) = 1$$
so:
$$v_0^2 = \frac{2R^2}{R + (4+ 2\sqrt3)y_0}$$

Answer 7

oh alright. I see that now

Answer 8

And you're cool with just having the formula? I mean, do you understand how to get that formula. I ask because formulas like these shouldn't be remembered because they're relatively easy to do. 10 different problems will probably have 10 different formulas just like that one. So don't just remember it as is because it's a solution of the current problem, that's all I'm saying.

Answer 9

Yeah that's what im having most trouble with is there's a lot of possible ways to solve one problem. Not so used to the idea, but knowing that such formula can be viable will help me be aware that I can use that for a specific situation. Thanks.

Answer 10

If it looks long, it's not, there's 6 lines of math in here. tops. the rest is just description of how and why and what for and how to notice these situations. Read it it will be helpful I promise (hopefully).

Well, that's kind of my point. I dont' know if this formula is like an official formula or just some half-baked thing I made. The point I was trying to make is that all these tasks are always done the same way. All the tasks on a slope are always done the same way, all the tasks with springs, or pendulums, or whatever are done the same way.
You always start with the same basic equations that start form some law that makes sense and is easy to remember. In our case it's the 4 or 5 equations in my first answer.

Because I did them sort of half baked first time around, I feel bad so I'll try to walk you through the standard procedure you use to solve all these kinds of tasks all the time. It's a bit longish but the mathematics is really really elementary.

First of all let's do a quick analysis on what we're dealing with. We have 2 motions in play here. One is the motion due to constant acceleration in the y direction. The second one is a constant motion without acceleration in the x direction.

This implies that in the x direction the velocity is constant and does not change with time. But because there is constant velocity the coordinates change with time. We will mark this by saying that coordinate x (distance) is a function of time and write it as x(t). Because velocity is constant it's the same as vx0 and is not a function of time.

In the y direction we have a constant acceleretion -> means it's not a function of time. But because there is an acceleration in play, the velocity must change with time. Because we have velocity, even if it's different in different times, the coordinate y (height) must change in time to. Therefore we have a, v(t), and y(t).

To derive it we start with the functions:
$$x(t) = v_0t \cos \theta\\
y(t) = y_0 + v_0 t \sin \theta - \frac{1}{2} g t^2$$
Why precisely these functions? The height in time, or y(t), is the formula for constant accelerated movement you probably did earlier this year. I'll skip showing how to get to that one for shortness but if you know how to - great! In any case, hopefully, you're familiar with it . The position x in time formula is just:
$$v = \frac{\Delta s}{\Delta t}\text{ but, remember the functions! }\\
\vec v = \frac{x(t)}{t}\\
x(t) = \vec vt$$
and then we remember that the movement in x direction is just dependant on the velocity in the x direction:
$$v_{x} = v_0 \cos\theta\text{ Put this in x(t)}\\
x(t) = v_0 \cos(\theta) t$$
Ok, now that that's sorted moving on. In these formulas we pick a time t and get to calculate the positions x(t) and y(t). But the problem is that we don't know the vx or the vy(t). We have to get rid of those somehow, or calcualte them, to be able to solve for x(t) and y(t).

We remember that we know x at some time t when the projectile hits the ground. That is we know the specific maximal x, so called range, when the projectile hits the ground. If the projectile is on the ground that means it's height, it's y coordinate, at that particular time is 0. This gives us a good motivation to put y=0 in the second formula.
$$- \frac{1} {2} g t^2 + v_0\sin(\theta) t + y_0 = 0$$
As it turns out the projectile is on the ground twice! Well, duh. Once we're launching it at t=0 and once it actually hits the ground. This will make things a bit more complicated because we can't just express t and put it back into x(t). Solve this for t:
$$t_{1,2} = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}\\
t_{1,2} = \frac{ -v_0\sin\theta \pm \sqrt{v_0^2\sin\theta^2 - 4\frac{-1}{2}gy_0} }{ 2\frac{-1}{2}g }\\
t_{1,2} = \frac{ v_0\sin\theta \pm \sqrt{v_0^2\sin\theta^2 + 2gy_0} }{ g }\\$$
Now we've got to chose which of the solutions to use. One of them will work out poorly because t will be 0, the second one will be once the projectile actually hits the ground. That time is obviously greater than 0. We check the square root to see if it could become a negative number and it can't. Everything in it will be positive (the v is positive, the sine of 15° is positive, the operation is plus, the initial height is positive and g is ~10 so positive) . Because the root will be positive we're not worried about imaginary number and conclude that the bigger, larger, of the two solutions for t will be the one with +.
$$t_2 = \frac{ v_0\sin\theta + \sqrt{v_0^2\sin\theta^2 + 2gy_0} }{ g }$$
Ok, what we just found is the time t that it will take the projectile to fall back to ground. That is, to say more precisely, we found a function t of initial velocity v0, angle of firing and initial height y0 of the time it takes the projectile to fall back onto ground.

We take that solution and put it back in the x(t) from the start of this. But we remember that the time we're putting in is from some time t we calculated that is also the time when the projectile hits the ground. Therefore we rename the x(t) into R, or D and call it range just to make sure not to confuse the two. Even if t looks like a function it's good to think about it like a number because we got it by setting y(t) = 0. So it's not exactly any t we want.
$$x(t) = v_0 t \cos \theta\\
R = v t_2 \cos\theta\\
R = v_0\cos\theta\frac{ v_0\sin\theta + \sqrt{v_0^2\sin\theta^2 + 2gy_0} }{ g }\\
\text{To make it less scary pull out the g: }\\
R = \frac {v_0 \cos \theta} {g} \left [ v_0\sin \theta + \sqrt{v_0^2 \sin^2 \theta + 2 g y_0} \right]$$

To make it usable in your scenario, we must flip this equation. We have to do it because the things we know are theta, g, y0 and R but we don't know v0. We want to calculate that because once we have it everything is much easier. So we flip it to get v0 out:
$$v_0^2 = \frac{R^2}{\cos\theta^2 \left(2*R\tan\theta + 2y_0\right)}$$
Now we put in the numbers we have and simplify as much as we can:
$$v_0^2 = \frac{2R^2}{R + (4+ 2\sqrt3)y_0}$$
as I have shown in my answers above.


Have I explained this good enough? Do you understand? If not, ask which part is confusing you. The math is long with a lot of symbols, but it's very very basic. 2 functions, a bit of quadratic equation solving, and some algebra to simplify. The thing I really want you to remember is the reasoning. We know how x(t) and y(t) look like. We observe what of the other variables in it we don't know. We find the one we can express over the other, in this case time t from y(t)=0, such that we can calculate one of the unknows from the first equation. You express the conditions you have through the second one, put them into first one, express your unknown variable and calculate it. If it looks long, it's not, there's 6 lines of math in here. tops. Your next task might give you measured times and you might have to express something else, but you will again be using the same procedure, just different letters.

Answer 11

I see! @ljetibo ! lots of thanks. sorry I'm just getting back to you right now, my week has been hectic in school. I like the explanation it helps as to why we're using the equations. Most of the people who have taught me kinematics just highlight what is given and start figuring out which formulas from there without really explaining why. It can be very misleading, so when it comes to me doing it on my own, I go really slow and blank out. Doing it with a professor makes it seem so easy but I can't manage to think about how to approach problems like these on my own.

I do have a couple questions, the y-coordinate you were talking about, that is the final height r which is why you got y = 0 because the ball is already on the ground?

And when you said, "This implies that in the x direction the velocity is constant and does not change with time. But because there is constant velocity the coordinates change with time. We will mark this by saying that coordinate x (distance) is a function of time and write it as x(t)," You're saying that the distance is not constant but the velocity and the x direction is not dependent on the time but the constant velocity that moves the projectile in a certain period of time?

You also said, "have a constant acceleration -> means it's not a function of time. But because there is an acceleration in play, the velocity must change with time," and the reason why that is because having a constant velocity in a period of time would make acceleration 0 but if the velocity increases/decreases constantly then so will acceleration?

Hope the questions made sense! I just want to make sure I'm understanding it right.

Answer 12

I guess best to take the questions one at a time:

QUESTION 1:
" I do have a couple questions, the y-coordinate you were talking about, that is the final height r which is why you got y = 0 because the ball is already on the ground? "

Yeah, the y coordinate is the height of the ball. It's a function of time - that means that it changes with time. It starts at 0 in the beginning and ends in zero at the end of the flight when it falls back to the ground. If you call that final time when the ball hits the ground with some letter - in example: T - then the final height can be written as y(T). At the start of the flight is, when we begin measuring our time, the height of the ball will be y(0) = 0. In general for any other time you write: y(t).

QUESTION 2:
""This implies that in the x direction the velocity is constant and does not change with time. But because there is constant velocity the coordinates change with time. We will mark this by saying that coordinate x (distance) is a function of time and write it as x(t)," You're saying that the distance is not constant but the velocity and the x direction is not dependent on the time but the constant velocity that moves the projectile in a certain period of time?"

I never noticed how bad my spelling can be. Well, the coordinate x is not a constant - for example when we throw the ball, time t=0, the position of the ball, x, will be zero. When the ball hits the ground, at time T, the position of the ball will be D. At some time in between x will be some distance less than the range D but bigger than zero. Because of that you say that x is a function of time.
However, velocity vx will be the same at *any* time t. This velocity will be vx right after you throw it, and it will be that same velocity vx just before it hits the ground. Velocity in the x direction *does not change with time*. So you say that velocity is a constant.
Whenever you have velocity involved you are immediately saying that body moves.
Saying "constant velocity" implies that its [body's] position is a function of time. It has speed, therefore it will move in time (v=s/t) - therefore position x is a function of time.
Saying "accelerated movement" implies that its position and velocity is a function of time. For example - acceleration of the force of gravity is 9.8 m/s^2 or 9.8 meters per second per second. That means that the body will move 9.8 meters per second faster(!) every second. Which is the same as saying that velocity is a function of time. But because we again have velocity involved - position must be a function of time.

In this case, you have one "constant velocity" movement - that in the x direction, and one accelerated movement - that in the y direction. This is because there are no forces acting in the horizontal direction - so by Newton's first law - there is no acceleration, but in the y direction there is the force of gravity and therefore the body will experience accelerated movement.

Is this what you asked?

QUESTION 3:
"You also said, "have a constant acceleration -> means it's not a function of time. But because there is an acceleration in play, the velocity must change with time," and the reason why that is because having a constant velocity in a period of time would make acceleration 0 but if the velocity increases/decreases constantly then so will acceleration?"

This ties into question 2. When I say "accelerated movement" I mean "constant accelerated motion" which means the acceleration doesn't change. That means acceleration the body experiences is the same at the start, in the middle and at the end of the flight.
The reason why the acceleration is constant is because you only have the force of gravity acting on the body. That force is constant and so the acceleration it "makes" the body feel is constant. When the acceleration is constant, then velocity is constantly **changing** - the velocity is a line upwards/downwards, and the position is a curve. When we're talking about constant acceleration that curve is quadratic.

http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/cacc.gif

But there is also nothing forbidding the acceleration from being a function of time. For example when you're in a car and you press the gas pedal - you start accelerating, but after a while because your can only has so much horse power, it can't accelerate any more. Then you've reached your top speed. In that case your acceleration is a curve that eventually ends up at zero. The motion equations in that case look different, but you do them again the same way.