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anonymous

  • 4 years ago

An object is thrown vertically upward at 18 m/s from a window and hits the ground 1.6 s later. What is the height of the window above the ground? (Air resistance is negligible.) a. 3.7 m d. 37 m b. 16 m e. 41 m c. 21 m

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  1. JamesJ
    • 4 years ago
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    You want to write down an equation for the position of the object at time t, call it y(t). y(t) = something involving initial velocity + something involving gravity. Then evaluate y(t=1.6 seconds) to find where the ground is.

  2. JamesJ
    • 4 years ago
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    Hint: if there were no gravity, then y(t) = 18t Now, what does gravity do to that equation?

  3. anonymous
    • 4 years ago
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    y(t)=18t-5t^2 plugin value for time t=1.6s y=16m

  4. anonymous
    • 4 years ago
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    ohhhhh!! is y(t) = Vi + g the equation?

  5. anonymous
    • 4 years ago
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    Why is y(t) = 18t - 5t^2? What is the general equation?

  6. anonymous
    • 4 years ago
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    heelp! D:

  7. anonymous
    • 4 years ago
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    http://ibphysicsstuff.wikidot.com/uniformaccmotion it is equation8 that u are looking fo ALL THESE ARE FOR UNIFORMLY ACCELERATED MOTION

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