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Dropping a ball and calculating its speed.

  1. Aug 6, 2014 #1
    1. The problem statement, all variables and given/known data
    If a person drops a 20kg ball from a height of 50m. What is the speed of the ball before it hits the ground? Neglecting air resistance.


    3. The attempt at a solution
    I'm not sure if this is a trick question. The formula I would use to solve this is final velocity = initial velocity + (at).

    I think that a would be 9.8 as its the speed of gravity but I don't have the time. Am I using the correct formula?
     
  2. jcsd
  3. Aug 6, 2014 #2

    Nathanael

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    Welcome to Physics Forums.

    So far, it's all good. But the tricky part of this question is supposed to be finding the time.


    First off, does it make sense that there is only one possible time that works? (Could there be more than 1 time that works?)

    Do you have any ideas about how you could find the time it takes a 20kg ball to fall a height of 50m?
     
  4. Aug 6, 2014 #3
    you don't have the time, as you said. Can you think of any kinematic equation that involves initial and final velocities but excludes the time taken?
     
  5. Aug 6, 2014 #4
    PE(initial)=KE(final) so you should see mass doesn't even affect anything.
     
  6. Aug 6, 2014 #5

    HallsofIvy

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    Not "speed" of gravity, acceleration due to gravity. If the acceleration is -9.8 m/s^2 then the speed after t seconds is v= 9.8t+ v0 (v0 is the initial speed and I am taking positive downward) and then the distance traveled is s= 4.8t^2+ v0t. Since the ball is "dropped" the initial speed is 0 so we have v= 9.8t and s= 4.8t^2.

    Since the ball falls 50m, solve 4.8t^2= 50 to find t.
     
  7. Aug 6, 2014 #6
    Use conservation of energy.It will be a lot easier.
     
  8. Aug 7, 2014 #7
    Is the formula to find velocity -> vf = vi + 2ad.

    vf = (0 m/s) + 2(9.81 m/s/s)(50 m)

    vf = 0 m/s + 981 m/s

    vf = 981 m/s

    vf = √(981 m/s) = 31.3 m/s

    Can someone confirm if the correct speed of the ball before it hits the ground is 31.3 m/s?
     
  9. Aug 7, 2014 #8

    ehild

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    It is wrong.

    vf is both 981 m/s and 31.3 m/s??????

    ehild
     
  10. Aug 7, 2014 #9
    Is the correct answer vf = 31.3 m/s?
     
    Last edited: Aug 7, 2014
  11. Aug 7, 2014 #10

    HallsofIvy

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    Yes, that is correct. At acceleration 9.8 m/s^2, the speed in t seconds is 9.8t m/s and the distance traveled is 4.9t^2. Since the ball fell 50 m you must have 4.9t^2= 50 so that [itex]t= \sqrt{50/4.9}[/itex] and then the speed at the end of the fall is [itex]9.8\sqrt{50/4.9}[/itex] which is approximately 31.3 m/s.

    As Satvik Pandey pointed out, you could also use "conservation of energy". Taking the ground as reference point, initially the ball has no kinetic energy because it is not moving but has potential energy, relative to the ground, mgh= m(9.8)(50)= 490m Joules. Just as it hits the ground it has no potential energy so it must have converted to kinetic energy: [itex](1/2)mv^2= 490m[/itex]. The "m" cancels and we have [itex]v^2= 980[/itex] and then [itex]v= \sqrt{980}= 31.3 m/s[/itex].
     
    Last edited: Aug 7, 2014
  12. Aug 7, 2014 #11
    Thank you all!
     
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