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Conservation of mechanical energy problem

  1. Apr 25, 2014 #1
    Problem: the masses of the javelin, discus, and shot are .8 kg, 2.0 kg, and 7.2 kg, respectively, and record throws in the track events using these objects are about 89 m, 69 m, and 21 m, respectively. Neglecting air resistance, (a) calculate the minimum initial kinetic energies that would produce these throws and (b) estimate the average force exerted on each object during the throw, assuming the force acts over a distance of 2 m. The answer is (a) 349 j, 676 j, 741 j (b) 175n , 338 n, 371n

    Relevant equations: KEi+PEi=KE+PEf
    Or 1/2mv^2+mgy= 1/2mv^2+mgy

    Attempt: I tried to use some kind of substitution to find v and y but I only had one equation so I couldn't use the substitution method.
  2. jcsd
  3. Apr 26, 2014 #2
    To have a projectile reach a given range, how can it be set in motion? Is there any freedom in choosing the parameters of the initial motion? How can that freedom be used to minimise the initial kinetic energy?
  4. Apr 26, 2014 #3

    Andrew Mason

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    For a given initial velocity, what is the launch angle that produces the longest range (longest horizontal distance of travel before hitting the level ground)? Use that angle when using the formula for initial velocity and range.
  5. Apr 26, 2014 #4
    The launch angle is not given.
  6. Apr 26, 2014 #5

    Andrew Mason

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    Write out the equation for the Range as a function of initial speed and angle. At what angle is the range maximum (i.e. keeping velocity constant)? If you use that value for the angle, then the equation provides the longest range for a given speed. They give you the range. Work out the speed.

  7. Apr 30, 2014 #6
    I used the equation ImageUploadedByPhysics Forums1398872549.449364.jpg . I used this to try to find the initial velocity and then used the initial velocity to try to find the kinetic energy when the potential energy is zero thus finding the energy needed for the system. I got the wrong answer. Is the formula I derived not correct?
  8. Apr 30, 2014 #7
    X represents the x distance at the top of the projectiles motion.
    Last edited: Apr 30, 2014
  9. Apr 30, 2014 #8
    Equation for trajectory:

    $$X = \frac{V_{0}^2\sin(2\theta)}{g}$$

    I think they want you to assume your angle ##\theta## is 45 degrees, because that is the angle at which a maximum trajectory will occur.

    You are given the trajectories but not the initial velocity, so that is what you want to solve the above equation for.

    After that, I am going to assume you know the equations for work and energy. Since they are asking for the average force over a given distance, the standard ##W = Fd## equation can be used.
  10. Apr 30, 2014 #9
    The way you wrote that equation is unclear. Could you send me a better representation?
  11. Apr 30, 2014 #10
    Umm, not sure how else to write it. If you were to read it out, you'd say that the trajectory of a projectile is equal to the initial velocity squared multiplied by the sine of 2 times the launch angle theta, divided by g, the acceleration of gravity.

    ##X## = trajectory
    ##V_{0}## = initial launch velocity
    ##\theta## = initial launch angle
    ##g## = acceleration of gravity
  12. May 1, 2014 #11
    Ok I get it. Thanks!
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