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What is the speed of the ball when it hits the ground?

  1. Oct 7, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown from the edge of a 10.0 m high cliff with a speed of 20.0 m/s at an angle of 30 degrees below horizontal. What is the speed of the ball when its hits the ground below the cliff?

    Options :
    1) 26.3 m/s
    2) 28.1 m/s
    3) 31.4 m/s
    4) 24.4 m/s
    5) 29.5 m/s

    I am new to the forum so bare with me I don't know how the forum works. I have attempted a solution for the problem. I would appreciate if you could just tell me the equation to work with.
    2. Relevant equations

    This is the equation I tried

    v = sqrt(2*g*h)


    3. The attempt at a solution

    v = sqrt(2*9.8*10)
    v = 14 m/s

    The answer is not even in the options. Clearly I am doing something wrong actually way off. May be I am mixing wrong concept here.

    I would appreciate explanation with equation.

    Thank you.
     
  2. jcsd
  3. Oct 7, 2015 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    That would be the speed if the ball were dropped from the height h. But it was thrown, so there will be some sort of addition of speeds...:smile:
     
  4. Oct 7, 2015 #3
    your ball also travels horizontally not just vertically try adding that to your equation
     
  5. Oct 7, 2015 #4
    So is it v=sqrt(2*9.8*10) +20
    so V = 14 m/s + 20 m/s

    v = 24 m/s?
     
  6. Oct 7, 2015 #5

    berkeman

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    Staff: Mentor

    14+20 is not 24...

    Besides, did you see the hint by Matejxx1? The velocity of the ball has horizontal and vertical components. Gravity affects the vertical component...
     
  7. Oct 7, 2015 #6
    v = sqrt (Vi^2 + 2*g*h)
    v = sqrt( 20^2 + 2*9.8*10)

    v = 24.4 m/s ?
     
  8. Oct 7, 2015 #7

    berkeman

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    Staff: Mentor

    That looks good. You could have also solved it by finding the initial Vx and Vy components, then added in the extra Vy component due to the acceleration of gravity over the 10m drop, and then taken the SQRT of the squares of the x & y velocity components. Maybe try that method to be sure you get the same answer. :smile:
     
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