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Thread: Mechanics question

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    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

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    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.
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    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

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    Last edited by JaxsenW; 17 Jul 2018 at 3:37 PM.
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    Re: Mechanics question

    Where did you get the initial velocity ?

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    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?

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    Re: Mechanics question

    Quote Originally Posted by JaxsenW View Post
    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



    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 for the horizontal and for the vertical or is it for both? Although I would assume it's the former since the question explicitly stated the coefficient was the same in both axes.
    Last edited by fan96; 19 Jul 2018 at 7:35 PM.

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    Re: Mechanics question

    If the resistive forces are and then we could write



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

    And differentiating the given equation should provide an easy solution for .

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



    which looks correct.

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    Re: Mechanics question

    Quote Originally Posted by fan96 View Post
    If the resistive forces are and then we could write



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

    And differentiating the given equation should provide an easy solution for .

    I tried graphing the motion of the projectile using my calculated value of 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!

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