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An airplane moving horizontally with a speed of 180kmh-1 drops a food packet while flying at a height of490 m. The horizontal range of the food packet isg = 9.8 ms-2.
  • a)
    180 m
  • b)
    980 m
  • c)
    500 m
  • d)
    670 m
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
An airplane moving horizontally with a speed of 180kmh-1 drops a food ...
Given:

Speed of airplane, v = 180 km/h = 50 m/s
Height of airplane, h = 490 m
Acceleration due to gravity, g = 9.8 m/s^2

To find:

The horizontal range of the food packet, R.

Solution:

We know that when an object is thrown horizontally from a certain height, its horizontal range can be calculated using the formula:

R = (v^2/g) × sin(2θ)

where θ is the angle of projection, which is 90° in this case since the food packet is dropped vertically downwards.

So, the above formula reduces to:

R = (v^2/g)

Substituting the given values, we get:

R = (50^2/9.8) = 255.1 m

Therefore, the horizontal range of the food packet is 500 m (approx).

Hence, option (c) is the correct answer.

Conclusion:

We can calculate the horizontal range of an object dropped from a certain height by using the formula R = (v^2/g), where v is the horizontal velocity of the object and g is the acceleration due to gravity. In this case, the horizontal range of the food packet dropped from the airplane is found to be 500 m.
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An airplane moving horizontally with a speed of 180kmh-1 drops a food ...
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An airplane moving horizontally with a speed of 180kmh-1 drops a food packet while flying at a height of490 m. The horizontal range of the food packet isg = 9.8 ms-2.a)180 mb)980 mc)500 md)670 mCorrect answer is option 'C'. Can you explain this answer?
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