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A helicopter is rising steadily upward at a rate of 3m per second .If an object is dropped from it ,then distance between the object and helicopter after 2 seconds ?
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A helicopter is rising steadily upward at a rate of 3m per second .If ...
Distance between the object and helicopter after 2 seconds when the helicopter is rising steadily upward at a rate of 3m per second

To determine the distance between the object and the helicopter after 2 seconds, we need to consider the motion of both the helicopter and the object independently.

Motion of the Helicopter:
The helicopter is rising steadily upward at a rate of 3m per second. This implies that its vertical velocity is constant at 3m/s.

Motion of the Object:
When the object is dropped from the helicopter, it experiences free fall due to gravity. The acceleration due to gravity is approximately 9.8m/s², acting in the downward direction.

Calculating the Distance:
To calculate the distance between the object and the helicopter after 2 seconds, we need to determine the vertical positions of both the object and the helicopter after this time.

The vertical position of the helicopter after 2 seconds can be calculated using the equation:
Distance = initial position + (velocity × time)
Since the helicopter is rising steadily upward at a rate of 3m/s, its initial position is 0m. Therefore,
Distance(helicopter) = 0 + (3 × 2) = 6m

The vertical position of the object dropped from the helicopter can be calculated using the equation for free fall:
Distance = initial position + (velocity × time) + (0.5 × acceleration × time²)
Since the object is dropped, its initial position is also 0m. Therefore,
Distance(object) = 0 + (0 × 2) + (0.5 × 9.8 × 2²) = 19.6m

Final Distance:
The distance between the object and the helicopter after 2 seconds is the difference between their vertical positions:
Final Distance = Distance(object) - Distance(helicopter)
Final Distance = 19.6m - 6m = 13.6m

Therefore, after 2 seconds, the object will be approximately 13.6 meters below the helicopter.
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