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anonymous

  • 4 years ago

A stone is thrown vertically upwards with a speed of 28 m/s. How fast is it moving when it reaches a height of 28.0 m? How much time is required to reach this height?

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  1. anonymous
    • 4 years ago
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    This is what I have so far: d = 28.0m, a = -9.8m/s(since its thrown upwards), initial velocity is 28.m/s. The unknown is the final velocity.For my equation, I figured i would use this one: vf^2 = vi^2+2g*d, where vf is final velocity, vi is initial velocity, g is gravity or acceleration and d is the distance My final answer for the first part was: vf^2=(28.0m/s)^2 + 2(-9.8m/s)(28m) -784m/s^2/-548.9 = 1.43

  2. phi
    • 4 years ago
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    how about v= 28 - g*t and distance d= 28t- (1/2) g *t^2

  3. anonymous
    • 4 years ago
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    so its d=(28m/s)t+1/2(-9.8m/s)t^2?

  4. phi
    • 4 years ago
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    with d=28 m solve for t to get the time. you get 2 answers (the first when the stone is going up, the second when the stone is falling)

  5. anonymous
    • 4 years ago
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    okay so: 20m=28m/st+1/2(-9.8m's^2)t^2 28m=-23.1 =1.21

  6. phi
    • 4 years ago
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    where did you get 20m? isn't it 28? you have to solve 4.9 t^2 -28t +28 = 0

  7. anonymous
    • 4 years ago
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    ops sorry, I meant 28.

  8. phi
    • 4 years ago
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    4.9 *1.21*1.21 -28*1.21+28= 1.29 which is not zero you must use t= (-b ± sqrt(b*b-4*a*c)/(2a) where a= 4.9, b= -28, c=28

  9. anonymous
    • 4 years ago
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    ah I see quadratic formula. Thanks.

  10. anonymous
    • 4 years ago
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    I got 4.42 and 1.29 for time, but the answers weren't right. I've plugged in all the values for the formula.

  11. phi
    • 4 years ago
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    I get t= 1.2922 seconds for when the stone first reaches a height of 28 m the velocity at this time is v= 28 - 9.8t= 15.336 m/s

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