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anonymous

  • one year ago

A rock is projected from ground level. Later, 4.0s after being projected, the rock is observed to strike the top of a 9.75m tall fence that is a horizontal distance of 240ft from the point of projection. Determine the speed with which the rock was projected.

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  1. Abhisar
    • one year ago
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    Hi! Have you tried solving it on your own first?

  2. Abhisar
    • one year ago
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    |dw:1440129172140:dw|

  3. Abhisar
    • one year ago
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    Suppose, the rock is projected with a velocity V, then the two rectangular components we are interested in are the horizontal and vertical components.

  4. anonymous
    • one year ago
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    ok

  5. Abhisar
    • one year ago
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    Let the vertical component be \(\sf V_v\) and horizontal component be \(\sf V_h\). We know that the rock travels a distance of 240 feets or 73.15 m in horizontal direction in 4 seconds. This is due to the horizontal component of its velocity. \(\sf \Rightarrow V_h=\frac{73.15}{4}\) Now, we also know that rock has traveled 9.75 m in vertically upward direction in 4 seconds. This must be due to the vertical component of its initial velocity. We can thus calculate the vertical velocity using second equation of motion. \(\Rightarrow \sf 9.75=V_v\times 4-\frac{1}{2}\times 9.8\times 4^2\)

  6. Abhisar
    • one year ago
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    Finally, after calculating the two components we can calculate its overall initial velocity by vectorally adding the two components. Remember that °since the components are rectangular, angle between them is 90°

  7. Abhisar
    • one year ago
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    Getting it?

  8. anonymous
    • one year ago
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    Yeah, somewhat.

  9. anonymous
    • one year ago
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    Could you please write what the second equation of motion is without substituting in the values?

  10. Abhisar
    • one year ago
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    \(\sf s=ut+\frac{1}{2}\times a \times t^2\) s= distance u=initial velocity t= time a=acceleration.

  11. anonymous
    • one year ago
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    So the acceleration value is the force of gravity?

  12. Abhisar
    • one year ago
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    Yes, in this case acceleration on the rock is in vertically downward direction and I have opted to chose downwards as negative. So a here = \(\sf -9.8 m/s^2\)

  13. Abhisar
    • one year ago
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    Note that there is no acceleration in horizontal direction hence we can use simply \(\sf Distance = speed \times Time\) for calculations in horizontal direction.

  14. anonymous
    • one year ago
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    Ok, so \[V_{V} = \frac{ 9.75 + 1/2 * 9.8 * 16 }{ 4}\] ?

  15. anonymous
    • one year ago
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    @Abhisar

  16. Abhisar
    • one year ago
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    Yes.

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