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anonymous

  • one year ago

A stone is thrown vertically upward with a speed of 24.0m/s (a) How fast is it moving when it reaches a height of 13.0m? b) How much time is required to reach this height? c) Why are there two answers to b?

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  1. anonymous
    • one year ago
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    I know there must be an equation to figure out answers a and b, but I don't see one in my notes. I know that the answer to c is because it goes up and goes back down so it passes 13.0m twice.

  2. IrishBoy123
    • one year ago
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    |dw:1441657866545:dw|

  3. anonymous
    • one year ago
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    For part A i plugged in what I have too the third equation and got 28.82, which is higher than the initial velocity, but wouldn't it be slowing down as it goes upward. Then I plugged what I have into equation 2 for part b., but one of the number I got was -2.24, which couldn't possibly be the answer.

  4. anonymous
    • one year ago
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    Consider the gravitational force as negative acceleration.

  5. anonymous
    • one year ago
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    First of all you need to find the time taken to reach the given height of 13m using the second equation, and then apply the time taken to the first equation respective to the initial velocity to find the reduced velocity by height=13m.

  6. anonymous
    • one year ago
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    The main reason for there being 2 answers is because you use the quadratic equation to solve this question which results in both positive and negative values as the x intercepts but fact of the matter is negatives cannot exist in a question like this.

  7. IrishBoy123
    • one year ago
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    check your numbers for part 1, 3rd equation is good way to go |dw:1441660508972:dw|

  8. IrishBoy123
    • one year ago
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    for the second part, use the 1st equation: \[v = u + at\] \[t = \frac{v-u}{a}\] you should only get one answer from this @Robert136 is bang on, if you use a quadratic you may expect to get several solutions

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