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Projectile motion of shotputter

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data
    A shotputter throws a shot with an initial speed of 15.5m/s at a 34.0 degree angle to the horizontal. Calculate the horizontal distance given that the shot leaves the athletes hand 2.2m above the ground???

    2. Relevant equations
    Distance=((initial velocity)^2 x sin (2xdegrees))/gravity(9.8)

    Although this is the level range formula used when height final = height of initial. How do you change this formula to accommodate the fact that the shot lands 2.2m lower than it started????????

    3. The attempt at a solution
    d=15.5^2 x sin 68/ 9.8 =29.09m, I know this must be wrong but i have read through the text book and it does not help me at all in finding how to work this out. I shows what to do if the degrees is not given but with the degrees what equation can i use??????????????
  2. jcsd
  3. Mar 21, 2010 #2
    There isn't any point by-hearting formulae such as the one above. Use the equations of motion for constant acceleration,
    v = u + at
    s = ut + at2/2
    v2 = u2 + 2as

    where u and v denote initial and final velocities, a denotes acceleration, and s denotes displacement.
  4. Mar 21, 2010 #3
    Yeah but none of them incorporate the angle of release given???? Im guessing thats an important variable, so shouldnt it be used
  5. Mar 21, 2010 #4


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    Homework Helper

    Initial height yo of the shot put is given.At the end it reaches the ground. So the final height y = 0
    y = yo + Vo*sinθ*t - 1/2*g*t^2.
    Solve for t. Then horizontal distance
    x = Vo*cosθ*t.
    Now solve the problem.
    Last edited: Mar 22, 2010
  6. Mar 21, 2010 #5
    Your equations must be wrong
    0=2.2+15.5*sin 34*t-1/2*9.8*t^2
    x=15.5*cos 34*-0.225
    this answer does not make sense
  7. Mar 22, 2010 #6


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    Homework Helper

    Negative time is not possible. While solving the quadratic equation, you will get two values of t. You have to select the positive value of t.
    Last edited: Mar 22, 2010
  8. Mar 22, 2010 #7
    alright, cheers
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