Projectile Motion: Path's length?

[babaliaris]

So I just learned about projectile motion. I understand why you can study it as two independent straight line motions . But this can give you a way to calculate total velocities or accelerations, just by adding its individual component of each vector.

If the initial position of the projectile is
$$
r = (x_0, y_0)
$$

then (1)

$$
v = v_xi + v_yj \\
a = a_xi + a_yj
$$

where (2)

$$
v_x = \frac{dx}{dt}, v_y = \frac{dy}{dt} \\
a_x = \frac{v_x}{dt}, a_y = \frac{v_y}{dt}
$$

SO (3)
$$
v_x = v_0, v_y = a*t \\
x-x_0 = v_0t, y-y_0 = \frac{1}{2}at^2
$$

Using the equations in (3) you can find the individual coefficients of the total v and a in (1) independently (well a is just -g here but anyways). But if you know the Δx and Δy of the two motions can you somehow find the length of the path that the projectile travels without knowing the function of the path itself?

 

[PeroK]

So I just learned about projectile motion. I understand why you can study it as two independent straight line motions . But this can give you a way to calculate total velocities or accelerations, just by adding its individual component of each vector.

If the initial position of the projectile is
$$
r = (x_0, y_0)
$$

then (1)

$$
v = v_xi + v_yj \\
a = a_xi + a_yj
$$

where (2)

$$
v_x = \frac{dx}{dt}, v_y = \frac{dy}{dt} \\
a_x = \frac{v_x}{dt}, a_y = \frac{v_y}{dt}
$$

SO (3)
$$
v_x = v_0, v_y = a*t \\
x-x_0 = v_0t, y-y_0 = \frac{1}{2}at^2
$$

Using the equations in (3) you can find the individual coefficients of the total v and a in (1) independently (well a is just -g here but anyways). But if you know the Δx and Δy of the two motions can you somehow find the length of the path that the projectile travels without knowing the function of the path itself?

Those equations define the path, so you don't have to calculate the path explicitly. That said, the path length is not easy to calculate.

 

[babaliaris]

Those equations define the path, so you don't have to calculate the path explicitly. That said, the path length is not easy to calculate.

I have not done this before but I will try to guess it:
$$
y = \frac{1}{2}at^2 + y_0\\
\\
x = v_0t \Leftrightarrow t = \frac{x}{v_0} \\
\\
f(x) = \frac{1}{2}a\frac{x^2}{v_0^2} + y_0\\
$$

This is the function of the path?

 

[PeroK]

I have not done this before but I will try to guess it:
$$
y = \frac{1}{2}at^2 + y_0\\
\\
x = v_0t \Leftrightarrow t = \frac{x}{v_0} \\
\\
f(x) = \frac{1}{2}a\frac{x^2}{v_0^2} + y_0\\
$$

This is the function of the path?

What happened to $v_{0y}$?

 

[babaliaris]

What happened to $v_{0y}$?

Isn't it zero? This is a projectile motion so I assumed v_0 only applies to the horizontal motion

$$
\vec{v_0} = v_0i + 0j
$$

 

[PeroK]

Isn't it zero? This is a projectile motion so I assumed v_0 only applies to the horizontal motion

$$
\vec{v_0} = v_0i + 0j
$$

It's not easy to get something off the ground unless you give it an initial upward component of velocity.

 

[babaliaris]

It's not easy to get something off the ground unless you give it an initial upward component of velocity.

Sorry I had in mind a plane flying at a velocity $v_0$ and at $t_0 = 0$ is dropping a bomb. Then everything below should work right?

$$
y = \frac{1}{2}at^2 + y_0\\
\\
x = v_0t \Leftrightarrow t = \frac{x}{v_0} \\
\\
f(x) = \frac{1}{2}a\frac{x^2}{v_0^2} + y_0\\
$$

 

[PeroK]

Sorry I had in mind a plane flying at a velocity $v_0$ and at $t_0 = 0$ is dropping a bomb. Then everything below should work right?

$$
y = \frac{1}{2}at^2 + y_0\\
\\
x = v_0t \Leftrightarrow t = \frac{x}{v_0} \\
\\
f(x) = \frac{1}{2}a\frac{x^2}{v_0^2} + y_0\\
$$

For that specific problem, yes. Note that you have also assumed that $x_0 = 0$.

Normally, for projectile motion, the only restriction is that $a_x = 0, a_y = -g$. If you have acceleration in the x-direction or resistance in either direction, then things get more complicated.

You should try the more general case where $v_{0y} \ne 0$.

Hint: introduce an initial angle $\theta$ (above the horizontal) at which the projectile is fired.

 

[babaliaris]

For that specific problem, yes. Note that you have also assumed that $x_0 = 0$.

Normally, for projectile motion, the only restriction is that $a_x = 0, a_y = -g$. If you have acceleration in the x-direction or resistance in either direction, then things get more complicated.

You should try the more general case where $v_{0y} \ne 0$.

Hint: introduce an initial angle $\theta$ (above the horizontal) at which the projectile is fired.

Thanks! I understood everything you said! But back to the original problem. How difficult it is too calculate the length of the path? Can we compare it $\frac{circle perimeter}{4}$ so if we know the radius of that circle we can calculate the length of that path? (Because the path of the projectile looks like it's an arc)

 

[PeroK]

Thanks! I understood everything you said! But back to the original problem. How difficult it is too calculate the length of the path? Can we compare it $\frac{circle perimeter}{4}$ so if we know the radius of that circle we can calculate the length of that path? (Because the path of the projectile looks like it's an arc)

The path is a parabola. You can look that up. Here, for example:

http://mathworld.wolfram.com/ParabolicSegment.html

PS The term you want is "arc length" of a parabola,

 

[PeroK]

Thanks! I understood everything you said! But back to the original problem. How difficult it is too calculate the length of the path? Can we compare it $\frac{circle perimeter}{4}$ so if we know the radius of that circle we can calculate the length of that path? (Because the path of the projectile looks like it's an arc)

Actually, here is another approach for your problem, although it is mathematically equivalent. We have:

$v^2 = v_x^2 + v_y^2 = v_0^2 +g^2t^2$

$v(t) = \sqrt{v_0^2 + g^2t^2}$

Where $v(t)$ is the speed of the projectile at time $t$.

Now, the distance traveled by the projectile is the integral of speed wrt time. So (an exercise for you), show that:
$$d(t) = \frac{v_0^2}{g} \int_0^{\frac{gt}{v_0}} \sqrt{1 + u^2} du$$
Where $d(t)$ is the distance traveled by the projectile, along its parabolic path, at time $t$.

 

[babaliaris]

Actually, here is another approach for your problem, although it is mathematically equivalent. We have:

$v^2 = v_x^2 + v_y^2 = v_0^2 +g^2t^2$

$v(t) = \sqrt{v_0^2 + g^2t^2}$

Where $v(t)$ is the speed of the projectile at time $t$.

Now, the distance traveled by the projectile is the integral of speed wrt time. So (an exercise for you), show that:
$$d(t) = \frac{v_0^2}{g} \int_0^{\frac{gt}{v_0}} \sqrt{1 + u^2} du$$
Where $d(t)$ is the distance traveled by the projectile, along its parabolic path, at time $t$.

$$
v = \frac{dx}{dt} \Leftrightarrow dx = vdt \Leftrightarrow \int_{x_0}^{x}dx = \int_{t_0}^{t}vdt \Leftrightarrow
x-x_0 = \int_{t_0}^{t}\sqrt{v_0^2 + g^2t^2}dt = \int_{t_0}^{t}\sqrt{v_0^2 (1 + \frac{1}{v_0}gt)^2}dt =|v_0|\int_{t_0}^{t}\sqrt{1 + (\frac{1}{v_0}gt)^2}dt
$$

Let $u = \frac{1}{v_0}gt$ then $du = \frac{1}{v_0}gdt$ and $u(t_0) = \frac{gt_0}{v_0}$ and $u(t) = \frac{gt}{v_0}$

$$
x_0 - x = |v_0|\frac{v_0}{g}\int_{t_0}^{t}\sqrt{1 + (\frac{1}{v_0}gt)^2} \cdot \frac{1}{v_0}gdt =
\frac{v_0^2}{g}\int_{u(t_0)}^{u(t)}\sqrt{1 + u^2} du
$$

I just can't understand how did you do this:
$v^2 = v_x^2 + v_y^2 = v_0^2 +g^2t^2$

$v(t) = \sqrt{v_0^2 + g^2t^2}$

 

[PeroK]

I just can't understand how did you do this:
$v^2 = v_x^2 + v_y^2 = v_0^2 +g^2t^2$

$v(t) = \sqrt{v_0^2 + g^2t^2}$

That's the usual relationship between the speed of a particle and the components of its velocity. Or, more generally, the magnitude of any vector and its components. For example:

$r = |\vec{r}| = \sqrt{x^2 + y^2 + z^2}$

Where $\vec{r} = (x, y, z)$. And:

$v = |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$

It's the same for any vector: force, momentum, electric field etc. It's the 3D generalisation of Pythagoras's theorem.

 

[babaliaris]

That's the usual relationship between the speed of a particle and the components of its velocity. Or, more generally, the magnitude of any vector and its components. For example:

$r = |\vec{r}| = \sqrt{x^2 + y^2 + z^2}$

Where $\vec{r} = (x, y, z)$. And:

$v = |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$

It's the same for any vector: force, momentum, electric field etc. It's the 3D generalisation of Pythagoras's theorem.

Well yes but I find something else:

$v_x = v_0$ and $v_y = -\frac{1}{2}gt^2$
$$
\vec{v} = v_x \cdot \vec{i} + v_y \cdot \vec{j} = v_0 \cdot \vec{i} - \frac{1}{2}gt^2 \cdot \vec{j} \\
||v|| = \sqrt{v_0^2 + \frac{1}{4}g^2t^4}
$$

LOL i think i just "burned" my mind after so many hours of reading

Well of course I used the wrong equation for $v_y$, $v_y = -gt$ so now I think i will find the correct one:
$$
\vec{v} = v_x \cdot \vec{i} + v_y \cdot \vec{j} = v_0 \cdot \vec{i} - gt \cdot \vec{j} \\
||v|| = \sqrt{v_0^2 + g^2t^2}
$$Can you tell me one more thing? Why do we need to integrate the magnitude of the velocity? I had this question a lot of times before but what does it mean to integrate a vector itself?
$$
\Delta{x} = \int_{t_0}^{t} (v_0 \cdot \vec{i} - gt \cdot \vec{j}) \cdot dt
$$

Is this something in mathematics that I might not know yet? Maybe does mathematics tell's us that "To integrate a vector function just integrate it's magnitude."? And if this is true why is that?

 

[Nugatory]

Can you tell me one more thing? Why do we need to integrate the magnitude of the velocity?

The magnitude $v(t)=|\vec{v}(t)|$ of the velocity vector is the speed. Thus $v(t)dt$ is the distance traveled in time $dt$. Integrating that gives us the total distance traveled.

Maybe does mathematics tell's us that "To integrate a vector function just integrate it's magnitude."? And if this is true why is that?

We aren't integrating a vector here, we're integrating an ordinary scalar function that happens to be the magnitude of a vector. (There is such a thing as integrating a vector function to get another vector, but that's not what we're doing here).

 

[A.T.]

Why do we need to integrate the magnitude of the velocity?

Because you want the path length.

...what does it mean to integrate a vector itself?

You can integrate the velocity components individually, to get the net displacements along each direction.

 

[PeroK]

$$
x_0 - x = |v_0|\frac{v_0}{g}\int_{t_0}^{t}\sqrt{1 + (\frac{1}{v_0}gt)^2} \cdot \frac{1}{v_0}gdt =
\frac{v_0^2}{g}\int_{u(t_0)}^{u(t)}\sqrt{1 + u^2} du
$$

One other point: we are not finding the difference in the x-coordinates here. We are integrating to find the arc length along the curve. I used $d(t)$ for the total distance traveled along the curve. The result is not $x-x_0$.

 

[Chestermiller]

$$
\Delta{x} = \int_{t_0}^{t} (v_0 \cdot \vec{i} - gt \cdot \vec{j}) \cdot dt
$$

Is this something in mathematics that I might not know yet? Maybe does mathematics tell's us that "To integrate a vector function just integrate it's magnitude."? And if this is true why is that?

This equation will give you the displacement vector between the initial point and the location at time t. This is the same as a secant drawn to the curve from the initial location to the location at time t. But you want the cumulative tangential distance along the path, which, of course, is greater than the length of the secant.

 

[Chestermiller]

See the current development in the Calculus and Beyond Forum, which provides the answer for the integral you are looking for: https://www.physicsforums.com/threads/length-of-polar-curve.968606/