1. Not finding help here? Sign up for a free 30min 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!

Find height, range and time of flight for projectile motion

  1. Oct 10, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-10-10_9-48-29.png

    2. Relevant equations
    kinematics.
    a=F/m

    3. The attempt at a solution
    100cos(70)=342
    100sin(70)=393.7
    X=1/2(-9.8)t^2+100t; t=0. t=20.4
     
  2. jcsd
  3. Oct 10, 2015 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You might want to check the arithmetic above before going further in this calculation. Also, don't forget to indicate units.

     
  4. Oct 10, 2015 #3

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Ok so you have an equation for the initial vertical velocity. What else do you know about the motion in the vertical plane? Final velocity at the top = ? Acceleration =? What other equations of motion do you know that might be relevant?
     
  5. Oct 10, 2015 #4
    y component = 94
    velocity at the top = 0
    acceleration from g = -9.8m/s

    I know my basic kinematics and a=f/m
     
  6. Oct 10, 2015 #5

    gneill

    User Avatar

    Staff: Mentor

    The problem states that the projectile is launched from the top of a building, yet doesn't state the height of that building. So what interpretation are we to place upon the Range and Time of Flight parts? Is a symbolic answer with an assumed starting height (say h) desired, or do we consider the landing elevation to be the same as the launch elevation? Is there any point to the building?
     
  7. Oct 10, 2015 #6
    I assume the range to be where the projectile lands (on the ground) and the time would stop at that point as well. I guess there is no point to the building although the elevation effects time and distance traveled.
     
  8. Oct 10, 2015 #7
    h=[v(initial)^2*sin(theta)^2]/2g = [(100)^2(sin70)^2]/-19.6 = -450.5

    Not sure how this can be negative unless it is accounting for the height of the building?
     
  9. Oct 10, 2015 #8

    gneill

    User Avatar

    Staff: Mentor

    The g in this case is just the constant g without sign. You might try deriving the expression for the maximum height to see why this is.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted