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Homework Help: Distance problem, Using Launcher need to find objects distance

  1. Nov 12, 2006 #1
    Hows it going guys, I need to now how to find distance of an object without actually shooting from my launcher. I have that it shoots 518.28 feet per sec ( initial velocity) ends at 0 ft/s (final velocity), that gravity's constant is -32.15 feet/sec2(9.8 meters/sec2), I also have that it shoots at a 45 degree angle and it launches 2.12 feet off the ground, objects weight is .0154 lbs. It feels like I have everything I need to find distance but I keep getting huge numbers. If you could solve this for me or just point me to the formulas that I would need I would greatly appreciate it:tongue: , This is going to help launch a projectile for a project and this formula will save me a lot of time and money
    Thanx again
  2. jcsd
  3. Nov 12, 2006 #2


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    Homework Helper

    So, you obviously used some formula to get these huge numbers, so, present your work. :cool:
  4. Nov 14, 2006 #3
    Here are the formulas
    ΔX= 1/2 (initial velocity in x direction + final velocity) Δtime
    vfx= vix+accelerationΔt
    Δx=viΔt+ 1/2 a (Δt)^2

    vf= final velocity
    vi=initial velocity
    a= acceleration
    Δ= displacement
    You must remember that in an equation vi and vf must be either along the X axis or along the Y axis.

    I solved it too
    Each equation can be applied to either the x or the y axis.
    that means
    Δx= vit+1/2at^2 is in the x direction
    Δy=vit+1/2at^2 is in the y direction

    Here it is solved
    vi in x direction = 518.28 cos 45
    vi in y direction = 518.28 sin 45

    -2.12= 366.48Δt-16t^2
    Solve for t
    t= 22.9 seconds
    t in y direction = t in x direction

    since acceleration = 0 in x direction
    Δx ~ 8392.392 ft

    That is a long distance. If it's way of, I'm sure there is a problem with you initial velocity. 518.28 ft/s is too much.
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