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Projectile Motion cannonball fired on top of a wall.

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data
    A knight fires a cannon ball from the top of the castle wall. the cannonball is fired at a speed of 55 m/s and an angle of 27 degrees above horizontal. A cannon ball is accidentally dropped and hits the moat below in 1.4s.

    how far from the wall does the cannon ball hit the ground?


    2. Relevant equations

    any of the kinematics.


    3. The attempt at a solution

    so i have figured by adapting a kinematic equation that the height of the wall is 9.61184 m tall. Yo= 1/2 ay(change in t)2

    i have placed my coordinates at the base of the wall.

    Xo= 0
    X= ???
    Vox= 55cos27
    Vx= 0
    aox=0
    ax = 0
    to = 0
    t = i believe this would be 3.9458 seconds, but i may be incorrect...


    Yo=9.61184 ( calculated by gravity acting on the ball that was dropped)
    Y=0
    Voy=55sin27
    Vy=0
    aoy= -g
    ay = -g
    to = 0
    t = i believe this would be 3.9458 seconds, but i may be incorrect...




    i know that my initial V in the x direction is 55cos27 and initial V in the y direction is 55sin27.

    now i have tried solving first for the time it takes for the ball to hit the ground in the Y direction. this is where i have trouble. i cant seem to isolate my change in time in any kinematic equation.

    i would assume once i have the change in time in the Y direction, i can then use that value to determine the final resting position in the X plane.....


    more or less i need help just understanding how to go about this problem in an organized systematic way.... any help at all would be greatly appreciated..
     
    Last edited: Sep 29, 2011
  2. jcsd
  3. Sep 29, 2011 #2
    You have a good start, but you need to calculate the height of the wall, this can be done using the ball that is dropped (presumably with initial velocity 0m/s). Once you have this you should check if your value of time is correct ( not quite sure how you calculated that) and if so then apply all these to find your answer!

    Thanks for laying the problem out clearly.
     
  4. Sep 29, 2011 #3


    my Yo would be the height of the wall.

    i will re calculate this to be sure, and then i guess use that value to determine the time it takes for the ball to travel up, hit its apex, and then reach the base of the wall, all in the y direction....

    if i take that time, can i use it with my kinematics to figure out where the ball will land in the x direction from the wall...?
     
  5. Sep 29, 2011 #4
    Hi yes,

    You have calculated this correctly, I had not noticed, my apologies. Once you have this the kinematic equation :
    [itex] S=Si+Vi \Delta t + \frac{1}{2} g \Delta t^{2}[/itex]
    can be used to solve the vertical motion of the projected ball, this will give you a quadratic to solve for the time taken for the canon ball to hit the ground (setting [itex] S=0 [/itex] ), once you have this time then you can solve for the horizontal distance.
     
  6. Sep 29, 2011 #5

    well ive now found my time using the quadratic formula. this is a step i had not envisioned doing so for starters, thank you for pointing me in this direction.


    now ive found my time, to be 5.45 seconds. this is the time it will take for the ball to travel straight up, and straight down to the ground due to the velocity in the y direction and the acceleration due to gravity. ( note i am using a negative acceleration with respect to where i have placed my coordinates.)


    i then used this travel time to calculate how far the ball will move in the X direction, due to the velocity in the X direction, with no acceleration, over the period of 5.45 seconds.

    using the same kinematic i used to get a quadratic. i then enter all of my X information.

    im coming out with an answer of 281.929m away from the base of the wall.

    i feel as though this is very close, but is still wrong......not sure where i am going wrong.

    i will have to go over this again to find my mistake,

    any suggestions as to where i might have miscalculated a number throwing off my final answer just a tad?


    again thank you for all the help, just being able to talk a problem like this through with someone else is enabling me to think about it in a different way.
     
  7. Sep 29, 2011 #6

    gneill

    User Avatar

    Staff: Mentor

    If you've got the correct x-direction velocity then you should be okay. d = v*t. Maybe a finger problem on the calculator?
     
  8. Sep 29, 2011 #7
    I agree with your value of the time t=5.45 secs from the quadratic, However using the same equation, the value of [itex] s_{inital}=0 [/itex] and [itex] a_{x}=0 [/itex]. So [itex] s=0+55cos(27) \Delta t + 0 = 267 m[/itex]

    So a little less than your values, but i wouldn't consider this to be to crazy since the ball is travelling at 55 m/s = 123 mph.

    It was good to talk through the problem, thanks.
     
  9. Sep 29, 2011 #8
    now this logically makes sense to me. d=v*t.

    but since we have a constant acceleration (0) i should get the same answer using kinematics.


    now, solving for the x direction....

    x= xo + Vox(Δt) + 1/2 ax (Δt)2


    in this case xo and ax go to 0, so for the sake of ease, i will drop them from my next step...


    solving....

    x= 55cos27 (5.45) + 1/2 (5.45)2

    x= 281.93



    but i know that the correct answer to the question is 270. ( we are looking for the answer in 2 sig figs. )


    where am i going wrong....why is this not working out using kinematics.....
     
  10. Sep 29, 2011 #9

    gneill

    User Avatar

    Staff: Mentor

    What's that 1/2 (5.45)2 term doing there? I thought you said there was no acceleration?
     
  11. Sep 29, 2011 #10
    ahh right. i cant drop it just because its 0 in this case.
     
    Last edited: Sep 29, 2011
  12. Sep 30, 2011 #11
    Hi

    Just to make sure you are clear: the term "[itex] 1/2 a_{x} \Delta t ^2 = 0 [/itex] if [itex] a_{x}=0 [/itex], as you have already stated.
     
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