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Shotputter - Projectile Motion

  1. Aug 6, 2006 #1
    "At a particular instant in time, a shot-putter pushes on a 5kg shot with a force of 90N at an angle of 45 degrees above the horizontal. If the only other force acting on the shot is due to gravity, what is the direction and magnitude of the shot's acceleration vector"

    Im stumped - i cant seem to grasp the concept of direction (in degrees) and how to work it out. No idea how the 5kgs comes into play either. Any hints to get me started would be fantastic :redface:
  2. jcsd
  3. Aug 6, 2006 #2
    If the magnitude of the acceleration is 90 ms^-2, and it acts at an angle of [tex] \pi /4 [/tex], then the magnitude of each component of acceleration (ignoring the gravity at this time, and using F = ma), is [tex] \frac{18}{\sqrt{2}} [/tex] (use trig to confirm this if you have to). Then, the acceleration due to gravity can be represented as the vector: 0i - gj. All you now have to do is combine the information I have just given you.
  4. Aug 10, 2006 #3
    Projectile Motion

    The accelerations acting on the projectile are (90/5) meters/(second^2) and g (acceleration due to gravity) acting downwards. Since acceleration is a vector quantity, the magnitude of the resultant will be a=(18^2 + g^2 + 2*18*g* cos(135)) ^1/2.

    The direction will be theta= tan^-1( 18*cos45/ (g-18*sin45) ) using tan(theta) = psin(alpha) / q+pcos(alpha) where theta is the angle made by the resultant with the base and alpha is the angle of projection.
  5. Aug 10, 2006 #4


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    Staff: Mentor

    Actually, I don't think gravitational acceleration comes into play until the shot leaves the person's hand. Think about it.
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