Projectile Motion problem involving air resistance

[Adrsya Rupam]

Homework Statement

Ok, so I am attempting to solve a projectile motion problem involving air resistance that requires me to find the total x-distance the projectile traverses before landing again.

Given:
$$\\<br /> m=0.7\text{kg}<br /> \\<br /> k=0.01 \frac{\text{kg}}{\text{m}}<br /> \\<br /> \theta=30 \degree$$

Homework Equations

$$F_{air}=kv^2$$

The Attempt at a Solution

I divided the dynamics of the problem into x-component and y-component equations:
From $$F_{x}=-kv_x^2$$, I got:
$$x\left(t\right)=\frac{m}{k}\ln \left(kv_it-m\sec \theta \right)$$

I divided the y-component of the motion into two parts--going up and going down:
I used $$F_{y1}=-mg-kv_y^2$$ for the going up part, and I used $$F_{y2}=-mg+kv_y^2$$
solving the differential equation for the going up part, I got:
$$v_y\left(t\right)=\sqrt{\frac{mg}{k}}\tan \left(\arctan \left(v_{yi}\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{g}{m}}\right)$$ -- which is where I am stuck on... because when I plugged in my given values, the graph doesn't look right as its t-intercept is greater than one which is found from solving this kinematically without air resistance.
solving the differential equation for $$v_y(t)$$ of the going down part and integrating it, I got:
$$y\left(t\right)=-\frac{m}{k}\ln \left(\cosh \left(t\sqrt{\frac{gk}{m}}\right)\right)+y_i$$

Can someone help me solve this problem?

 

[ehild]

Homework Statement

Ok, so I am attempting to solve a projectile motion problem involving air resistance that requires me to find the total x-distance the projectile traverses before landing again.

Given:
$$\\<br /> m=0.7\text{kg}<br /> \\<br /> k=0.01 \frac{\text{kg}}{\text{m}}<br /> \\<br /> \theta=30 \degree$$

Homework Equations

$$F_{air}=kv^2$$

The Attempt at a Solution

I divided the dynamics of the problem into x-component and y-component equations:
From $$F_{x}=-kv_x^2$$, I got:
$$x\left(t\right)=\frac{m}{k}\ln \left(kv_it-m\sec \theta \right)$$

I divided the y-component of the motion into two parts--going up and going down:
I used $$F_{y1}=-mg-kv_y^2$$ for the going up part, and I used $$F_{y2}=-mg+kv_y^2$$solving the differential equation for the going up part, I got:
$$v_y\left(t\right)=\sqrt{\frac{mg}{k}}\tan \left(\arctan \left(v_{yi}\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{g}{m}}\right)$$ -- which is where I am stuck on... because when I plugged in my given values, the graph doesn't look right as its t-intercept is greater than one which is found from solving this kinematically without air resistance.
solving the differential equation for $$v_y(t)$$ of the going down part and integrating it, I got:
$$y\left(t\right)=-\frac{m}{k}\ln \left(\cosh \left(t\sqrt{\frac{gk}{m}}\right)\right)+y_i$$

Can someone help me solve this problem?

No, this is not correct. The force of air resistance is opposite to the velocity vector and its magnitude is proportional to v². v² is a scalar, it does not have components.

 

[Ray Vickson]

Homework Statement

Ok, so I am attempting to solve a projectile motion problem involving air resistance that requires me to find the total x-distance the projectile traverses before landing again.

Given:
$$\\<br /> m=0.7\text{kg}<br /> \\<br /> k=0.01 \frac{\text{kg}}{\text{m}}<br /> \\<br /> \theta=30 \degree$$

You have not given an initial speed, so your problem is not completely specified.

For any given initial speed you can set up and solve the DEs numerically, and I think that is about the best you can do (in view of the criticism of "ehild" in #2).

Perhaps, though, you can get usable approximations by attempting something like a perturbation theory approach in which you essentially expand in powers of the small parameter $k$.

 

[Adrsya Rupam]

No, this is not correct. The force of air resistance is opposite to the velocity vector and its magnitude is proportional to v². v² is a scalar, it does not have components.

You have not given an initial speed, so your problem is not completely specified.

For any given initial speed you can set up and solve the DEs numerically, and I think that is about the best you can do (in view of the criticism of "ehild" in #2).

Perhaps, though, you can get usable approximations by attempting something like a perturbation theory approach in which you essentially expand in powers of the small parameter $k$.

Sorry, I forgot to mention that-- the initial velocity is 9 m/s

 

[Adrsya Rupam]

No, this is not correct. The force of air resistance is opposite to the velocity vector and its magnitude is proportional to v². v² is a scalar, it does not have components.

What do you mean? Can't any force in vectorspace be divided into
[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/18d95a7845e4e16ffb7e18ab37a208d0ab18e0e0, https://wikimedia.org/api/rest_v1/media/math/render/svg/3dc8de3d8ea01304329ef9518fad7a6d196c4c01 components?

 

[Ray Vickson]

What do you mean? Can't any force in vectorspace be divided into
[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/18d95a7845e4e16ffb7e18ab37a208d0ab18e0e0, https://wikimedia.org/api/rest_v1/media/math/render/svg/3dc8de3d8ea01304329ef9518fad7a6d196c4c01 components?

The friction force acts along the negative of the tangent to the trajectory. See, eg.,
http://wps.aw.com/wps/media/objects/877/898586/topics/topic01.pdf
or
http://young.physics.ucsc.edu/115/range.pdf

The case of $\vec{f}_{\text{friction}} = - k \vec{v}$ is tractable, but not your case of $\vec{f}_{\text{friction}} = - k |v|^2\, \vec{v}/|v| = -k |v| \vec{v}$.

 

[Adrsya Rupam]

The friction force acts along the negative of the tangent to the trajectory. See, eg.,
http://wps.aw.com/wps/media/objects/877/898586/topics/topic01.pdf
or
http://young.physics.ucsc.edu/115/range.pdf

The case of $\vec{f}_{\text{friction}} = - k \vec{v}$ is tractable, but not your case of $\vec{f}_{\text{friction}} = - k |v|^2\, \vec{v}/|v| = -k |v| \vec{v}$.

Ahh, I get it... would I need to use numerical approximation methods?

 

[Ray Vickson]

Ahh, I get it... would I need to use numerical approximation methods?

Yes. That is what I said in #3.