1. ## Mechanics question

Hey guys, we have an assignment due first day back next term, and I'm just slightly confused with part of it.

This is the question

A projectile is fired from a gun at 45 degrees to the horizon. The displacement of the projectile in the x axis is defined by the following function:

X=50ln(t+0.25)-50ln(0.25)

The coefficient of resistance is the same in both the x and y axes, and the resistance is proportional to the velocity squared.

From this, I need to find the coefficient of resistance as part of one of the questions, and I am confused how to do this.

Thanks so much

Sent from my SM-G935F using Tapatalk

2. ## Re: Mechanics question

Hey ill give you some hints. Try to determine an expression for velocity and then use the info given in the question to equate it with resistance. note the importance of the proportional bit. Then resolve the vectors to determine the force equation, noting this is projectile motion etc. Then since you're given the coeff of resistence is the same in the horizontal and vertical, you should be able to make some progress from there.

3. ## Re: Mechanics question

Thanks heaps!
I've gotten this far but this is the part in struggling with...
I just dont know where to go from here?
Do I just equate those or...?

Thanks again

Edit: sorry, mechanics is the topic I am struggling with most atm. Thanks again

Sent from my SM-G935F using Tapatalk

4. ## Re: Mechanics question

Where did you get the initial velocity $200\sqrt2$ ?

5. ## Re: Mechanics question

Just realised I should've used 200 not 200sqrt2
But I got it from taking dx/dt and subbing in t=0 to get 50/0.25=200 initially
Is the process right?

Sent from my SM-G935F using Tapatalk

6. ## Re: Mechanics question

Originally Posted by JaxsenW
Just realised I should've used 200 not 200sqrt2
But I got it from taking dx/dt and subbing in t=0 to get 50/0.25=200 initially
Is the process right?

Sent from my SM-G935F using Tapatalk
I'm not sure, but I noticed on your fifth line you use

$\ddot x = \frac{d}{dx}\left(\frac12v^2\right)$

which, unless I've read the question wrong, isn't necessarily true since the overall velocity of the projectile is different from its horizontal and vertical velocities.

Also you've differentiated velocity to get acceleration but then immediately integrate back again.

And just to clarify when the question says that horizontal/vertical resistance is proportional to the velocity, does that mean the overall velocity or the respective horizontal/vertical velocities?

i.e. is the resistive force $k\dot x ^2$ for the horizontal and $k\dot y ^2$ for the vertical or is it $kv^2$ for both? Although I would assume it's the former since the question explicitly stated the coefficient was the same in both axes.

7. ## Re: Mechanics question

If the resistive forces are $k\dot x^2$ and $k\dot y^2$ then we could write

$\ddot x = -k\dot x^2$

(as the only horizontal force acting on the projectile AFTER launch is the resistance)

And differentiating the given equation $x = 50 \log(4t+1)$ should provide an easy solution for $k$.

I tried graphing the motion of the projectile using my calculated value of $k$ and got this:

which looks correct.

8. ## Re: Mechanics question

Originally Posted by fan96
If the resistive forces are $k\dot x^2$ and $k\dot y^2$ then we could write

$\ddot x = -k\dot x^2$

(as the only horizontal force acting on the projectile AFTER launch is the resistance)

And differentiating the given equation $x = 50 \log(4t+1)$ should provide an easy solution for $k$.

I tried graphing the motion of the projectile using my calculated value of $k$ and got this:

which looks correct.
Thankyou!
Yes I confirmed with my teacher that k is proportional in the x and y axis respectively, so this works.
Thanks so much for your help!

Sent from my SM-G935F using Tapatalk

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•