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Projectile question with no time or velocity or range

  1. Jan 19, 2015 #1
    1. The problem statement, all variables and given/known data

    A fly ball just clears a 10.0 meter fence on the way down at a 45 degree angle, and strikes the ground 8.0 meters beyond the fence. Calculate the speed of the ball when it left his bat (at ground level).

    since it lands at ground level dy=0
    a=9.81m/s2 since it is the only acceleration acting on a projectile.

    2. Relevant equations

    dy=vit+1/2at2

    v=dx/t


    3. The attempt at a solution
    Ok so ive been all over the place with this question and i have no idea how to solve it. There is no mass so i cannot use any type of energy formula, no time which means i will at some point have to find it, no velocities to start off with, and no overall range or dx distance.
    If i had the dx distance then i would be able to solve it since i would have 3 of 4 necessary info for the dy=vit+1/2at2 equation, but with out it im lost.
     
  2. jcsd
  3. Jan 19, 2015 #2

    lightgrav

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    be careful to use enough adjectives to indicate what portion of the trajectory your equation is about.
    For example, you DO know some distance, 10m ... and another one, 8m ... (one of them will be used as a negative).
    Draw this, so you know which is which.
    It might be better to use dy = vy,f t - ½ a t2 ... this t is only after clearing the fence.
     
  4. Jan 19, 2015 #3

    I set up a triangle which has the 10.0 meter fence as the y axis and the 8.0 meter distance as the x axis and the balls path is the hypotenuse in between. I do have those distances but without any form of time i still cant calculate anything from that triangle i dont think. And the Vy,f is the final velocity correct? which means that it would simply be zero, leaving me with 0 = -4.9t2
     
  5. Jan 19, 2015 #4

    gneill

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    Staff: Mentor

    The ball's path won't be a straight line: It follows a parabolic trajectory due to the influence of gravity.

    You've recognized the problem within the problem, namely the trajectory from the top of the wall to the ground:

    Fig1.gif

    If you assume some speed for the ball at the top of the wall, say vw, and the projectile is moving at 45° below the horizontal from the top of the wall, and you know that it strikes the ground at 8 m from the wall, can you determine the speed vw?

    You should be able to write equations of motion for the x and y directions, and write the initial velocities in the x and y directions in terms of vw (sin(45°) and cos(45°) have simple values). There's enough information to find vw.

    Once you know the speed vw at the top of the wall you should be able to find the impact speed at the ground.
     
  6. Jan 19, 2015 #5

    Im not sure if i understand. I know that you can find the components (such as Vx and Vy) for an angled velocity by using the equations Vx=Vcosθ and Vy=Vsinθ. Im doing this for an advanced physics assignment at school and im the only one doing it and ive been running over all of my notes and i still cant find anything to help my case.
     
  7. Jan 19, 2015 #6

    gneill

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    Staff: Mentor

    Write the kinematic equations for the x and y components of the motion from the top of the wall. Assuming an initial speed vw, what are the x and y positions of the object with respect to time?
     
  8. Jan 19, 2015 #7
    Ive used the Vx=Vcosθ and Vy=Vsinθ ( where θ = 45) yet all i get from both of those is 0.7071V. You mention in respect to time and i assume that i am misinterpreting what you mean to convey, as there is no time as of right now in my equations and problem.
     
  9. Jan 19, 2015 #8

    gneill

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    Staff: Mentor

    Write the kinematic equations of the motion! y(t) = ...., x(t) = ....

    Yes, you are introducing a time variable t, but you will soon get rid of it by combining the equations. So, just start by writing the equations of motion. You want to solve for the initial velocity at the top of the wall that results in the projectile striking the ground 8 m from the base of the wall. You have enough information given: the launch angle, initial height, and horizontal distance at the point of impact.

    By the way, you should memorize the numerical value for sin(45°) and cos(45°) so you don't have to drag the trig functions around through all your calculations. Other convenient sin and cos values to know are those for 30° ;)
     
  10. Jan 19, 2015 #9


    God i feel like such an idiot, ive got you helping me and im still confused. Ok so these are the equations ive got,
    d = vo • t + 0.5 • a • t2
    vf = vo + a • t
    vf2 = vo 2 + 2 • a • d
    d = (vo + vf)/ 2 • t
    and the way you put it in your post ("y(t) = ...., x(t) = ....") confuses me as i dont understand that format of the equation; which truth be told only makes me feel worse about myself.
    I kind of understand what you mean by combining the two equations to cancel out the time variable, but again i cant move forward untill i understand what equation to use; and i promise i am decently smart i just lack abstract thinking, i need to do or see something once before i can understand it or repeat it.
     
  11. Jan 19, 2015 #10

    gneill

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    Staff: Mentor

    y(t) means y as a function of t, x(t) means x as a function of t.

    From the assumed initial speed and given launch angle you can write the initial x and y velocity components. Then you can write the equations for the x position and y position with respect to time using those initial velocities. Those are the basic projectile motion equations.

    So, write the equation for the y position given the initial y-position, initial y-velocity and the known acceleration due to gravity. Also write the x-position given the initial x-position and x-velocity (assume that the wall is at x position 0 m).
     
  12. Jan 19, 2015 #11

    lightgrav

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    math notation writes "f, as a function of x" like f(x).
    physics folks sometimes use math notation, but more often use subscripts, like your vf ... final is a time, after all.
    Almost always, we use x as horizontal; for vertical we either use y or z.

    Notice that your "d" equation has an acceleration ... if the acceleration is NOT zero, which direction is it? (x or y) ... call it that (dx or dy)
    If the acceleration IS zero, which direction is it? (x or y) ... call it that (dx , or dy)
    Now, solve the simpler equation for "t" ... and plug it (that symbolic formula) into the nastier equation.
     
  13. Jan 19, 2015 #12
    ok that makes some sense to me, so the y(t) = ...., x(t) = .... gneill was talking about refers to dx =...and dy =.... equations? such as dy = vi • t + 0.5 • a • t2
    And i am assuming that the "nastier equation" is referring to the formula that i create to cancel out a variable that i don't have yet?

    The d equation with acceleration is the dy (vertical) as it is gravity that acts upon it. and the one without acceleration (=0) is horizontal.
     
  14. Jan 19, 2015 #13

    lightgrav

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    yeah, go for it! by coincidence, in this problem the vx = -vy , only as the ball crosses the fence
    ... for the vertical one vfence,y =/= vground,y
     
  15. Jan 19, 2015 #14
    Ok well ill give it a shot tomorrow and post back on here if ive gotten it or not.
     
  16. Jan 20, 2015 #15
    Ok so i understand what you mean about vx=-vy, so with this being said i should be able to take dx/t = the rearranged formula for dy=vit+1/2at2, my only issue is that it doesnt work out. im assuming that im rearranging something wrong or my logic is flawed.
     
    Last edited: Jan 20, 2015
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