A watermelon cannon fires a watermelon vertically up into the air at a velocity of +11.5 m/s, starting from an initial position 1.20 meters above the ground. When the watermelon reaches the peak of its flight, what is (a) its velocity, (b) its acceleration, (c) the elapsed time, and (d) its height above the ground?
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The acceleration would be -9.81m/s^2, the acceleration due to gravity,
which is pulling down on the object at all times. Although the
velocity would be zero at the peak of its flight, this does not negate
the effects of gravity upon the watermelon.
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The acceleration would be -9.81m/s^2, the acceleration due to gravity, which is pulling down on the object at all times. Although the velocity would be zero at the peak of its flight, this does not negate the effects of gravity upon the watermelon.
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