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Projectile Motion

  1. Dec 10, 2011 #1
    How to find the initial velocity required for a projectile to be launched at an angle theta, with a distance 'd' and height 'h'.

    An expression is required. I am unable to find one using all three.

    Thanks in advance.
  2. jcsd
  3. Dec 10, 2011 #2


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    What do you mean by height? --- the maximum height achieved by the projectile?

    In my opinion, it's generally best to solve such problems by considering that motions in the vertical & horizontal directions are independent, rather than looking for some "grand" solution for each possible scenario.
  4. Dec 10, 2011 #3
    Yeah, I mean maximum height achieved by the projectile. It's a spring that needs to be stretched 'x' so that it attains a certain initial velocity 'v' so that it hits a target 'd' meters away and 'h' meters high from the same level of launch.

    I know that x = √(mv2/k)

    I know the value of 'k'. I need an equation for the value of 'v'.

    Any ideas?
  5. Dec 10, 2011 #4


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    Let's see if I understand. The target is d meters away, horizontally, and is h meters higher in elevation than the launch position. You want to know what velocity, v, is needed to launch the projectile at an angle of θ and have the projectile hit the target. Is that correct?
  6. Dec 10, 2011 #5


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  7. Dec 10, 2011 #6
    Yes that is correct. Given height, distance and angle of launch, find velocity.

    PS: That is an awesome drawing!
  8. Dec 11, 2011 #7
    I'd start off by writing newtons law for the projectile

    since the force is gravity we have F=mg and so for the y axis

    you can either solve this differential equation or look up the solution, it should be in any intro physics textbook, and you'll find that the y position of the particle is;
    [itex]y=a_0+a_1t+\frac{g}{2} t^2[/itex]
    and for the x position you'll find;

    you'd then set [itex]a_0 = b_0 = 0[/itex] to make the projectile be launched from the origin of co-ordinates
    you set the velocity as V Sin and V Cos, which gives you;
    [itex]y=V Cos( \theta ) t + \frac{g}{2} t^2[/itex]
    [itex]x=V Sin( \theta )t[/itex]

    so you now have to equations which you can solve, given some time you want to hit the target at, for V and theta
    [itex]Tan( \theta )=\frac{x}{y-\frac{g}{2}t^2}[/itex]
    [itex]\theta = arctan(\frac{x}{y-\frac{g}{2}t^2})[/itex]
    where x and y are the x and y of the target

    and you can substitute your theta back in to find the V

    if you already had the angle theta (I couldn't tell from your OP) you can put it in the tan equation and solve for t, then sub that into the x equation to get v

    good luck!
  9. Dec 11, 2011 #8
    I have the angle theta, the distance the target is away and the height of the target. I don't have the time.

    I need to find the initial velocity of the projectile.

    From your equations,

    I get the "v sin" and "v cos" equations for x and y.
    Shouldn't the 'x' eqn use v cos(theta) and 'y' be v sin(theta)?

    But where do I get the value of 't'?

    Also, the tan(theta) doesn't give me the velocity?
  10. Dec 11, 2011 #9


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    Write the equations for the horizontal and vertical components of the motion. Use v cos(θ) for the initial horizontal velocity, and v sin(θ) for the initial vertical velocity. When the projectile is at the target, how will these equations look (what values are in place for the x and y variables)?
  11. Dec 11, 2011 #10
    OK this is my equation for 't',

    t=[itex]\sqrt{}((xtan\vartheta - y)/(-0.5g))[/itex]

    Can I substitute this for 't' in the equation,

    v = x/(cos[itex]\vartheta[/itex]t)
  12. Dec 11, 2011 #11


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    Can you show how you arrived at your equation for t?
  13. Dec 11, 2011 #12
    x = vcos[itex]\vartheta[/itex].t

    t = x/(vcos[itex]\vartheta[/itex])


    Isolating 't',

    t2 = (x tan[itex]\vartheta[/itex] - y)/-0.5g

    t = [itex]\sqrt{}((x tan\vartheta[/itex] - y)/-0.5g)
  14. Dec 11, 2011 #13
    Also, when we talk about 'x', is it the total distance travelled by the projectile or is it just half the distance (that is the horizontal distance covered when it has reached its maximum height)?
  15. Dec 11, 2011 #14


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    Where does this come from?
  16. Dec 11, 2011 #15


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    It's whatever you want it to mean :smile: It's the value of x that corresponds to the corresponding value of y at the same instant of time.

    So if the time of interest is the instant that the projectile is at the target, then x is the horizontal distance from the launch point to the target, and y is the vertical height of the target w.r.t. the launch point.
  17. Dec 11, 2011 #16
    All I did is switch it to ,
    [itex]y=V Sin( \theta ) t + \frac{g}{2} t^2[/itex]
    [itex]x=V Cos( \theta )t[/itex]

    and then divided y/x to get tanθ and hence the equation.
  18. Dec 11, 2011 #17


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    Okay, well this is really going the long way around!

    [itex] y = v\;sin(\theta)\;t + \frac{1}{2} g t^2 [/itex] (for some reason choosing a negative value for the constant g :confused:)
    [itex] x = v\;cos(\theta)\;t[/itex]

    [itex] sin(\theta) = \frac{y - \frac{1}{2} g t^2}{v t}[/itex] and

    [itex] cos(\theta) = \frac{x}{v t} [/itex]

    [itex] tan(\theta) = \frac{y - \frac{1}{2} g t^2}{x} [/itex]

    Solving for t:

    [itex] t = \sqrt{\frac{2}{g} (y - x\;tan(\theta)} [/itex]

    The reason I say that this is the long way around is that you can find a value for t very easily from the equation for the horizontal motion. Substitute that into the equation for the vertical motion and solve for v.
    Last edited: Dec 11, 2011
  19. Dec 11, 2011 #18
    Now, I'm confused. Let me explain the whole scenario.

    I have to make a spring launcher that will fire a stretched spring. I will be given a target that is 'x' meters away and 'y' meters high w.r.t launch point. I will also be given an angle of launch theta.

    My solution (as of now),

    x = v cos θ .t -----------------------------------------------------------------Eqn 1
    y = v sin θ + 0.5gt2 (where g is -9.8, I will keep that in mind!)

    We get the equation,

    tan θ = (y-0.5gt2)/x

    Solving for t,

    t = √((x tanθ - y)/ (-0.5g)) (again g is -ve)

    Now from eqn 1,

    v = x / (cos θ.t)

    Substituting for t,

    v = (x) / [ cosθ . √((x tanθ - y)/(-0.5g)) ]

    Now to find the stretch x,
    k -> Spring constant, z-> stretch

    Elastic Energy in spring = Kinetic energy

    0.5kz2 = 0.5mv2

    z = √([mv2 ] / k)

    Substituting for v,

    z = √[ (m/k)( (x2) / (cos2θ ((x tanθ - y)/(-0.5g)) ) ]

    That was my grand equation. Let me know what you think.
  20. Dec 11, 2011 #19


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    I think you'll find that:

    [itex] t = \sqrt{\frac{2}{g}(y - x tan(\theta)}[/itex]

    Is the spring compressed before or after the projectile is loaded onto it? The reason I ask is that there will be gravitational PE involvement in the compression/expansion of the spring.
  21. Dec 11, 2011 #20
    It is stretched before launch so that it can launch itself with the energy stored in it.
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