Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Projectile Fired At A 45 Degree Angle

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data
    A projectile is fired at a 45 degree angle and its just barely able to make it above a 6m high fence that is 100 meters away. What was the projectile's initial velocity?


    2. Relevant equations
    x and y components?


    3. The attempt at a solution
    Well the problem with this one is that I don't know where to start. I mean this is basically the reverse of everything I've learnt about projectile motion. I think I should be calculating the x and y components first but I'm not even sure how to do that.
     
  2. jcsd
  3. Oct 4, 2011 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In all these constant acceleration kinematics problems, the best place to start is by writing two lists. The first is a list of everything that you know already. The second, is a list of the things you want to know. You should do this for each component (vertical and horizontal) of the motion.
     
  4. Oct 4, 2011 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    step 1. Write down all of the equations you know that might be useful here.
     
  5. Oct 4, 2011 #4
    1. v=u+at (cannot use because there is no time or final velocity)
    2. s= (u+v)/2 x t (cannot use because no time, distance or inital and final velocity)
    3. s= ut + 1/2 at^2 (cannot use because there is no initial or final velocities or time or distance)
    4. v^2 = u^2 +2as (cannot use this one because we have no inital or terminal velocity or distance)

    Others
    s=d/t (I'll most likely (definately) use this at the end when I have my horizontal and veritcle components)

    I'm sorry, but I honestly dont see how I can just have 1 variable when using any of these kinematics (when solving for the horizontal and veritcle components). Could using simultaneous equations be answer?

    What I already know:
    Distance from fence to launch point = 100m
    Gravitational Acceleration = -9.8m/s or 9.8m/s
    Angle of launch = 45 degrees
    the x and y components would be equal? :confused:

    What I need to know:
    x and y components
    total distace covered while in the air
    time of flight?
    inital velocity
     
    Last edited: Oct 4, 2011
  6. Oct 4, 2011 #5

    NascentOxygen

    User Avatar

    Staff: Mentor

    It's fired with a velocity V at 45 degrees. Write the horizontal component of firing speed, in terms of V. Write the vertical component of firing speed, in terms of V.
     
  7. Oct 4, 2011 #6
    Sorry mate, but I'm not exactly sure what is meant by "in terms of" :confused:
     
  8. Oct 4, 2011 #7

    NascentOxygen

    User Avatar

    Staff: Mentor

    Okay. :smile:

    It's fired with a velocity V at 45 degrees. Write the horizontal component of that.

    Now write the vertical component.
     
  9. Oct 4, 2011 #8
    So the verticle component would be

    Sinθ = o/V
    Sin(45 = o/V


    The horizontal component

    Cosθ = a/V
    Cos(45 = a/V

    where hypotenues is equal to V

    what now? this is where I get stuck because I don't know if I have enough infomation to move on or am I missing something:confused:
     
  10. Oct 4, 2011 #9

    NascentOxygen

    User Avatar

    Staff: Mentor

    What is o? What is a? This topic reserves a for acceleration. If you use the same symbols for different things, you will soon get confused.

    Besides, I can't see your expression for the horizontal velocity.
     
  11. Oct 4, 2011 #10
    sorry, I was using trig functions where o = opposite and a = adjacent.

    I might just keep it as: a = adj and o = opp

    Can I use Cosθ = adj/hyp or speed=distance/time to figure out the horizontal component?
     
  12. Oct 4, 2011 #11
    if the initial speed is v0, the horizontal component is u0=v0*cosθ and the vertical w0=v0*sinθ.

    Do you see this? :)
     
    Last edited: Oct 4, 2011
  13. Oct 4, 2011 #12
    I think so, but I don't see how you can get numbers out of those letters :biggrin: :confused:
     
  14. Oct 4, 2011 #13
    one step at a time ;)
    you may be used to other notations like v_x instead of u or whatever, but get comfortable with what components mean first.
     
  15. Oct 4, 2011 #14
    this excerise is a bit tricky, cause as you say, it's "the other way around". but we know, considering the equations for the two components, that;

    6m= - gt^2+V0*sinθ*t (the relevant velocity is here the vertical hence the sine)

    100m= V0*cosθ*t (the horizontal velocity calls for the use of a cosine)


    everybody concur? :)
    Since we have 2 equations and 2 unknowns, finding the answer for V0, the initial velocity, is just some algebraic puzzlework.
     
    Last edited: Oct 4, 2011
  16. Oct 4, 2011 #15

    NascentOxygen

    User Avatar

    Staff: Mentor

    Yes, it does look like 3 unknowns, but since we were told that theta is 45 degrees, then that leaves only two unknowns.
     
  17. Oct 4, 2011 #16
    You're absolutely right, I forgot to insert for theta. It's best to keep the symbols as long as possible though, to obtain a more general solution :)
     
  18. Oct 4, 2011 #17
    So simultaneous equations?
     
  19. Oct 4, 2011 #18

    NascentOxygen

    User Avatar

    Staff: Mentor

    Yes.
     
  20. Oct 5, 2011 #19
    I'm not sure if this is correct, but what if I say that the launch point was 6 metres above ground and I removed the fence. Would the time of flight be that same if the launch was on the ground and the fence was 100m away?
     
  21. Oct 5, 2011 #20
    if the velocity of the y component is zero at 100 meters away and Vy=V0 - gt, then what does that tell you about the time it takes to get there?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?