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Ranya99

  • one year ago

A watermelon cannon fires a watermelon vertically up into the air at a velocity of +9.5m/s m/s, starting from an initial position 1.2 meters above the ground. When the watermelon reaches the peak of its flight, what is: (a) its velocity (b) its acceleration (c) the elapsed (d) the height above the ground

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  1. anonymous
    • one year ago
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    a) Here , at the top of the flight , the velocity is zero

  2. anonymous
    • one year ago
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    b) as the acceleration of the watermelon is same at all time in the air, acceleration = -g = -9.8 m.s^2

  3. anonymous
    • one year ago
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    c) time taken to reach top is t Using first equation of motion v = u + a*t 0 = -9.5 + 9.8 * t t = 0.97 s the time taken to reach the top is 0.97 s

  4. anonymous
    • one year ago
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    d) height above the ground = v^2/2g + Ho height above the ground = 9.5^2/*2 * 9.8 + 1.2 height above the ground = 5.81 m the height above the ground is 5.81 m

  5. Ranya99
    • one year ago
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    Thanks a Lot

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