Laly asked in Science & MathematicsPhysics · 2 years ago

S.O.S!!!!!!!?

A flowerpot falls from a windowsill 15.0 m above the sidewalk.

a. How fast is the flowerpot moving when it strikes the ground?

b. How much time does a passerby on the sidewalk below have to move out of the way before the flowerpot hits the ground?

2 Answers

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  • 2 years ago

    As the flowerpot falls 15.0 meters, it has an acceleration of 9.8 m/s^2. Let’s use the following equation to determine its velocity when it strikes the ground.

    vf^2 = vi^2 + 2 * a * d, vi = 0

    vf^2 = 2 * 9.8 * 15 = 294

    vf = √294

    This is approximately 17.1 m/s.

    b. How much time does a passerby on the sidewalk below have to move out of the way before the flowerpot hits the ground?

    Let’s use the following equation to determine the time for this to happen.

    vf = vi + a * t, vi = 0

    √294 = 9.8 * t

    t = √294 ÷ 9.8

    This is approximately 1.75 seconds. I hope this is helpful for you.

  • oubaas
    Lv 7
    2 years ago

    final speed V = √2gh = √19.612*15 = 17.15 m/sec

    falling time t = √2h/g = √30/9.806 = 1.75 sec

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