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anonymous

  • one year ago

A ball is dropped from the top of a tower 100m high simultaneously another ball is thorwn upward with a speed of 50 meter per second. After what time do they cross each other? (a) 1s (b) 2s (c) 3s (4) 4s

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  1. anonymous
    • one year ago
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    When they cross each other they will have the same displacement. Set up equations for the displacements of each and equate them.

  2. anonymous
    • one year ago
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    but which equation, i've tried by appling S=vi t+1/2 at^2 but that does not work for me.Please can you solve it?

  3. anonymous
    • one year ago
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    You are on the right track. For the ball dropped downward, vi = 0. So\[\color{green}{\vec{d} = \frac{1}{2}\vec{g}t^2 }\] Set up the other equation and equate the two and solve for t.

  4. anonymous
    • one year ago
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    i've tried many methods but couldn't solve it,please solve it if you can

  5. Pawanyadav
    • one year ago
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    Assume displacemt x and 100-x for thrown and release one. Then 100-x=0.5gt^2 x=100-5t^2 Again For the thrown ball x=50t-5t^2 As time for both balls are equal Equating both values you will get the answer 2.

  6. shamim
    • one year ago
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    Did u get it?

  7. shamim
    • one year ago
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    We wanna help u. Response plz!

  8. anonymous
    • one year ago
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    ok,thanks i've got it

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