Find the time when the projectile runs into the hill (with air resistance)

[Moolisa]

Homework Statement

Consider the projectile motion shown in the figure below, where air resistance can be approximated by the linear regime. The red line represents an inclined hill, which can be assumed a straight line passing through the origin, making an angle with the horizontal

(a) Find an equation that defines the time t[SUB]h[/SUB] when the projectile runs into the hill. You can use all the results derived in lecture for projectile motion with linear drag.

(b)Accurate to first order in the linear drag coefficient, solve for the time t[SUB]h[/SUB]

Relevant Equations

v(y)=v(ter) + (v(y0) -v(ter)e^(-t(h)/τ) where tau=m/b EQ 1

v^y=v_ter + (v_y0 -v_ter) e^-t_h/τ where tau=m/b EQ 1

Okay, for part a, I used Eq 1
I let v_y=v_y(t_h)=0 --->The reasoning is that the projectile would stop moving for a short time when it hits the incline, but I have a feeling that reasoning is faulty

I let v_y0=v₀sinθ

Then the equation became

(v_ter-v₀sinθ)/v_ter= e^-t_h/τ

I multiplied both sides by the natural log, then got

t_h= (m/b)-v₀sinθ/g

I think this is wrong, but I'm not sure how to approach it differently or where I'm going wrong.

Obviously, I have not attempted part b, but I'm exactly sure I know what it is asking me to do. What does accurate to the first order linear drag coefficient mean?

 

[Antarres]

Your reasoning that you should search for a point where velocity is zero is definitely not good, since that brief moment you speak of is not described by the trajectory of the projectile(That trajectory judging by it's equation would continue through the inclined object if it was possible, you just stop it there since you know there is going to be a collision).

However, I find it very hard to read the equation you've given, and also, is that equation just for vertical component of velocity? In case when there is air resistance, both horizontal and vertical components will change over time, since drag force will have components along both axes.

So it would be good if you provided formulas for both components. That is in the case where you are not yet able to solve differential equations from which those formulas would be derived(This is listed as introductory physics problem, so I wouldn't know your level of math). In case you're familiar with differential equations, the first step would be to solve Newton's second law for both components of velocity, then find the trajectory, and finally find the intersection of the trajectory with this line that represents the inclined hill.

 

[callieferg]

The first step will be looking at the vertical motion of the projectile. You need to find the time where the velocity in the y direction becomes zero - if you double that, that will give you the time it takes for the projectile to reach the ground (if the incline weren't there).

Also, you need to separate the initial velocity into its separate components first. It's easier if you look at the equations of motion of the projectile separately for x and y.

Another formula that might help you:
$m$ $dv_y/dt = -b(v_y - v_{ter})$

To clarify, the key would be integrating the equations of motion to find the positions in the x and y directions.

 

[Antarres]

The first step will be looking at the vertical motion of the projectile. You need to find the time where the velocity in the y direction becomes zero - if you double that, that will give you the time it takes for the projectile to reach the ground (if the incline weren't there).

For motion with air drag, the time of ascension and descension isn't equal since air drag switches signs between ascending and descending, so it is not symmetrical with respect to the highest point in the trajectory.

 

[haruspex]

You need to find the time where the velocity in the y direction becomes zero - if you double that, that will give you the time it takes for the projectile to reach the ground (if the incline weren't there).

As @Antarres has pointed out, the motion is not symmetric about the apex. Further, it would not be helpful to find the time to return to y=0; and thirdly, it is generally a mistake to break trajectories into before apex and after apex: as long as the same equations apply in both, that's just extra work.

you need to separate the initial velocity into its separate components

That only works because it is linear drag, of course.

Another formula that might help you

From post #1, it appears that solving the acceleration equation for 2D linear drag is something already covered in class and Moolisa has the result for velocities. Whether @Moolisa also has the result for displacements (which is needed here) is unclear.

 

[Moolisa]

Thank you, I solved the problem and got the correct answer. It was tedious... Thanks for the help, I'll edit it to say it's been solved

 

[Moolisa]

Thanks everyone, I was able to solve it