An airplane moving horizontally with a speed of $180 k m h^{- 1}$ drops a food packet while flying at a height of $490 m$ . The horizontal range of the food packet is $g = 9.8 m s^{- 2}$ .

180 m

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980 m

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500 m

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670 m

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Solution

The correct option is B $500 m$
$u = 180 k m h^{- 1} = 50 m s^{- 1}$
Horizontal range $= u \sqrt{\frac{2 h}{g}} = 50 \sqrt{\frac{2 \times 490}{9.8}} = 500 m$ .

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An airplane moving horizontally with a speed of 180 kmh−1 drops a food packet while flying at a height of 490 m. The horizontal range of the food packet is g=9.8 ms−2.

The correct option is B 500mu=180 kmh−1=50 ms−1Horizontal range =u√2hg=50√2×4909.8=500 m.

An airplane moving horizontally with a speed of $180 k m h^{- 1}$ drops a food packet while flying at a height of $490 m$ . The horizontal range of the food packet is $g = 9.8 m s^{- 2}$ .

The correct option is B $500 m$
$u = 180 k m h^{- 1} = 50 m s^{- 1}$
Horizontal range $= u \sqrt{\frac{2 h}{g}} = 50 \sqrt{\frac{2 \times 490}{9.8}} = 500 m$ .