1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

DISTANCE travelled by the projectile

  1. Jul 20, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the DISTANCE travelled by the projectile?
    I know how to calculate the displacement, but this includes both x and y components, so how do I calculate the distance? I'm sure it must include both height and vertical displacement components



    2. Relevant equations



    3. The attempt at a solution

    I have no clue really. Everything I try to do, it just gives me the displacement. I just don't know how to get the 'h' factor in? Do I use pythagaros? Because it doesn't apply for a curved surface.
     

    Attached Files:

  2. jcsd
  3. Jul 20, 2013 #2

    Zondrina

    User Avatar
    Homework Helper

    Wait are you trying to find the distance the object flew in total or are you trying to find its horizontal displacement ( I'm sure it's displacement though ).

    With that in mind, to get the horizontal displacement of an object use :

    $$\vec{Δd}_H = \vec{v}_H Δt = \frac{ \vec{v}_R^2 sin(2θ) }{ \vec{a} }$$

    depending on which information you have been given or have been able to retrieve.
     
    Last edited: Jul 20, 2013
  4. Jul 20, 2013 #3

    CAF123

    User Avatar
    Gold Member

    What is ##\vec{v}_R?## That equation does not make sense. You cannot square a vector or divide by one.
    @judas_priest: Distinguish between horizontal and vertical motion. Write the equation for the displacement of the body in each direction. Then find the horizontal displacement in terms of known quantities by eliminating variables between the two equations.
     
  5. Jul 20, 2013 #4

    Zondrina

    User Avatar
    Homework Helper

    ##\vec{v}_R## is the resultant velocity ( Hypotenuse ).

    EDIT : Yeah you're right, It should be :

    $$\frac{v_R^2 sin(2 \theta)}{a}$$

    without the vector quantities.
     
  6. Jul 20, 2013 #5

    Curious3141

    User Avatar
    Homework Helper

    Most elementary projectile mechanics question limit themselves to asking about the maximum horizontal distance travelled by the projectile, also known as the range. This is very easy to work out, and the well-known formula for a projectile launched on a flat surface has already been quoted. For a projectile launched from a height, the answer is quite simple to work out from first principles.

    But if the question is actually asking for the total distance travelled by the projectile on its curved path, it's not impossible. Tedious, but not impossible. The distance is simply ##\displaystyle \int_0^T v(t)dt##, where ##T## is the total time of flight, and ##v(t)## is the instantaneous speed of the projectile expressed as a function of time.

    Using the resultant of the horizontal and vertical components of velocity at time ##t##, it is possible to get an expression for ##v(t)##. Integrating that is a little tough, and a hyperbolic trig substitution is probably the quickest way. And simplifying the resulting mess is daunting. But an answer can be derived fairly quickly.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted