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Shot putter physics problem

  1. May 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that a shot putter can put a shot at the worldclass speed v0 = 14.00 m/s and at a height of 2.160 m. What horizontal distance would the shot travel if the launch angle θ0 is (a) 45.00° and (b) 41.00°? The answers indicate that the angle of 45°, which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

    2. Relevant equations
    Horizontal distance = 14 * cos θ * [(14 * sin θ ÷ 9.8) + (2.106 ÷ 4.9)^0.5]


    3. The attempt at a solution
    so i have been using the formula above and am not getting the right answer. I don't know what to do. wrong formula maybe?
     
  2. jcsd
  3. May 30, 2012 #2

    cepheid

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    Rather than just trying to blindly apply formulae, let's think through the problem. In order to figure out the horizontal distance travelled, you need to figure out how much time the projectile spends in the air (so that you can multiply the horizontal speed by this time to get the distance).

    So, how much time does the projectile spend in the air? It is the sum of two different times: 1) The time required to go up to the maximum height, and , 2) the time required to come back down to the ground from this maximum height.

    How would you go about finding 1 and 2, knowing that the vertical acceleration is constant (it is due to gravity)?

    As hinted at in the problem, the slightly tricky part here is that the starting height is 2.106 m, whereas the final height reached (upon hitting the ground on the way back down) is 0.00 m.
     
  4. May 31, 2012 #3
    so i tried to use d=1/2gt^2 and got .664 seconds for time up but when i time required to come back down d for final height is 0 so how can i do 0=1/2gt^2 without getting 0?
     
  5. May 31, 2012 #4

    cepheid

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    You can't use d = 1/2gt^2 on the way up, because you don't know what d is. I.e. you don't know what maximum height above 2.106 m it reaches.

    What you can do is to use what you know about speed vs time for constant acceleration. What *must* be true about the vertical speed of the object at the instant that it reaches its max height?
     
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