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Physics test-Projectile Motion shotputter

  1. Sep 10, 2011 #1
    1. The problem statement, all variables and given/known data
    My Physics teacher told us that there is going to be a question on the test where we have to find the initial velocity of a shot put. He said that we will know the length of the shot putters arm and the launch angle. I was wondering what i methods there are to solving this question.


    2. Relevant equations
    S=ut +0.5at^2
    S=vt-0.5at^2
    v^2=u^2 +2as
    S=0.5(v+u)t
    v=u+at
     
  2. jcsd
  3. Sep 10, 2011 #2
    what's the use of length of shot-putter's arm??
     
  4. Sep 10, 2011 #3
    you need either height or range of projectile along with angle to calculate initial velocity.
     
  5. Sep 10, 2011 #4
    i think the shot putters arm length will be used as a displacement so at t=0s s=0m and at say t=0.6s s=0.9m.. Im not sure because we havent been told the question exactly just what the basics about it so im looking for possible ways to solve it
     
  6. Sep 10, 2011 #5
    when the ball is still in shot-putter's arm it still is getting velocity.you then might need acceleration provided to object by arm.But then it's question of newton's law of motion not projectile.
     
  7. Sep 10, 2011 #6
    yeah i dont know its gonna be a tough test i think but i want to top the class lolso i have to work as hard as possible and make sure there isnt a question they can ask me which i wont know
     
  8. Sep 10, 2011 #7
    There is something you always need to remember when dealing with problems where there is no air resistance involved and with a shotput no drag is an excellent assumption. The one constant of the problem is that the horizontal component of velocity remains constant throughout the flight of the projectile.

    So if given an initial angle of inclination of flight, you can easily compute the horizontal velocity component (a constant) by trigonometry. Then you only need to compute the time of flight which is based on the vertical component of velocity, Vv. The vertical component of velocity will have the same magnitude going up as it does coming down for any respective height above release point. So the time of the flight is twice the time it takes to get to its maximum elevation or twice the time it takes to fall to the point of release achieving a vertical velocity Vv, the vertical component of the initial velocity. So if Vv is the vertical component of initial veloicity at some angle, the time of flight is t = 2*Vv/g, g being the gravitational acceleration. The distance the shotput travels is then distance = Vh * t, where Vh is the horizontal component of velocity based on the initial angle of inclination.

    Another simple fact that you may already know. If you fire a bullet horizontally and at the same instant drop a bullet from the same elevation, both hit the ground at the same time.

    Good luck with your test. The problem could always be stated in reverse order so what I might do is do a practice problem with some initial velocity at some angle. Determine the distance. Then using the distance, work the problem backwards to get either the initial velocity or angle of inclination. Needless to say, when going in reverse, you must come up with the same initial conditions.
     
  9. Sep 10, 2011 #8
    If we are assuming the shotputter's is not so short that his height is negligible then the time of flight would be twice the time to max height + the time to fall the remaining height to the ground.

    The only way I can think you would be able to solve this problem without knowing how high the shotput is launched from is by:
    i) Using the time the shotput leaves his neck as t=0
    ii) Use the length of his arm as s
    iii) Try to find out the velocity when the shotput leaves his hand when his arm is fully extended.

    I have nothing to attempt to work this out on so I'm not entirely sure if it can be done but it's worth giving it ago.
     
  10. Sep 11, 2011 #9
    You could have any 4 pieces of information and might be asked to supply the fifth. Because there is no drag (aerodynamics involved), the flight is symmetric about the apex (except for path below the release point). A good example of the flight path when air drag is included is watching PGA golf on TV. Ball spin also has an effect there. The ball comes down at a larger angle than what it was when released. But that is a more complicated problem that is beyond the scope of a first course in physics.

    1. Initial velocity
    2. Total distance traveled horizontally
    3. Angle of inclination
    4. Height above ground of release point
    5. Distance traveled downrange at apex of flight

    You could make up a problem supplying 4 of the 5 items above. Then compute the fifth. With the answer you get for the fifth and three others, see if you can compute any one of the others that's missing. Obviously, you should get the same number each time as you go through the combinations.

    Four of the five items are not always needed. For instance suppose your teacher provides #5, #1, #4, #2 then asks for #3. Items 4 and 2 are not needed.
     
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