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Calculate initial speed of an athlete - projectile motion

  1. Mar 30, 2016 #1
    1. The problem statement, all variables and given/known data
    An athlete can cover 3.0 m for a standing long jump. Assuming that this athlete jumps at an angle of 45 degrees to the horizontal. calculate the speed at which the jumper can launch herself from a standing start.
    Assume g =10 ms^-2
    and no air resistance
    Assume that horizontal velocity never changes
    2. Relevant equations
    v=u+a*t
    x=(u+v/2)*t
    x=u*t+(a/2)*t^2
    x=v*t-(a/2)*t^2
    v^2=u^2+2*a*x
    v= final velocity
    x= displacement
    a= acceleration
    u= initial velocity
    t= time
    sin theta*initial velocity= vertical initial velocity
    cos theta*initial velocity= horizontal initial velocity
    3. The attempt at a solution
    In my scientific calculator I observed that cos 45 = sin 45 and thus the initial vertical and horizontal velocity must be equal to each other. From my observations, I do not think I have enough information to find this from the formula's listed. I don't really know what to do from here =/
     
  2. jcsd
  3. Mar 30, 2016 #2

    haruspex

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    There are two dimensions to be considered. You can apply those SUVAT equations separately in each.
    In general, a SUVAT equation involves four variables of initial speed, final speed, acceleration, distance covered, and time taken. So given any three you can find the other two, i.e. there are effectively six unknowns altogether forthe whole motion.
    In this case, you know the distance and acceleration in each direction. That settles four of them. You also know the initial velocities are the same in each direction, so that's one more equation. What else is the same for both motions?
     
  4. Mar 30, 2016 #3

    Merlin3189

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    So can you produce an equation for the horizontal component and another for the vertical component, then equate them to find another piece of data?
     
  5. Mar 30, 2016 #4
    How do we know the acceleration for the horizontal component? And what do you mean by 'initial velocities are the same in each direction, so that's one more equation' ?
     
  6. Mar 30, 2016 #5

    Merlin3189

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    If horizontal velocity never changes, what is horizontal acceleration?

    UH = UV is a simple, but valid equation.
     
  7. Mar 30, 2016 #6

    Simon Bridge

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    note: sin(45deg) = cos(45deg) = 1/√2
    so if the initial speed is u, the vertical and horizontal components are u/√2

    Newton's Laws ... what are the forces acting on the jumper?
    Alternately - if you are assuming the horizontal velocity does not change, then you can just use the definition of acceleration.

    He means the components of the initial velocity:
    The initial velocity would be ##\vec v(0)=v_x(0)\hat\imath + v_y(0)\hat\jmath## but it is easier to work in components.
    At time t>0, ##\vec v(t)=v_x(t)\hat\imath + v_y(t)\hat\jmath## ... this just says you can always break the velocity into horizontal and vertical components.
    ##v_x(0)=v_y(0)## is an equation that says "the components of the initial velocity are the same as each other" - maths is a language.

    You are going to end up solving simultanious equations - so just write down all the relationships you know that apply to the situation.
    i.e. if it takes time T to make the jump, then x(0)=0 and x(T)=3m ... y(0)=0 what is y(T)?
     
  8. Mar 30, 2016 #7
    How do I know the displacement vertically? and I don't know what else is the same in terms of both motions. They both have the same initial velocity but the difference in acceleration (which for horizontal it must be 0, thank you) implies that the motions must be different.
     
  9. Mar 30, 2016 #8

    haruspex

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    Is the landing point higher than the take-off point? Lower?
    Think through the five variables I listed for each of the two directions, horizontal and vertical. Which ones are the same in both directions?
     
  10. Mar 30, 2016 #9
    Time? Time can't be different for either directions of motion, the direction has no effect on it whatsoever.
     
  11. Mar 30, 2016 #10

    haruspex

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    That's right, the two motions occur over the same period of time. So what equations can you now write?
     
  12. Mar 30, 2016 #11
    x=u*t+(a/2)*t^2
     
  13. Mar 30, 2016 #12

    haruspex

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    Is that for the horizontal or the vertical direction? You need one equation for each, using distinct symbols where appropriate.
     
  14. Mar 30, 2016 #13
    Could you use it for both?
     
  15. Mar 30, 2016 #14

    haruspex

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    You can use the same generic equation, but the variables may take different values, so use a distinct name for each variable that might have a different value.
     
  16. Mar 30, 2016 #15
    Ok so I have two equations
    Vertical component:
    0=1/√2*u*t-5*t^2
    Horizontal component:
    3=1/√2u*t

    From this I can rearrange the horizontal equation to get t by itself and then substitute it into the vertical equation to attain u?
     
  17. Mar 30, 2016 #16

    haruspex

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    Yes.
     
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