A ball is dropped from a height of $19.6 m$ . The distance covered by it in the last second is

19.6

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14.7 m

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4.8 m

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9.8 m

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Solution

The correct option is C $14.7 m$
Let x be the distance covered in the last second. Now using $v = u + g t$ , we get
$\sqrt{2 g h} = \sqrt{2 g (h - x)} + g \sqrt{2 g (19.6)} = \sqrt{2 g (19.6 - x)} + g$
Solving the above equation we get,
$x = 14.7$ m

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A ball is dropped from a height of 19.6m. The distance covered by it in the last second is

The correct option is C 14.7mLet x be the distance covered in the last second. Now using v=u+gt, we get √2gh=√2g(h−x)+g√2g(19.6)=√2g(19.6−x)+gSolving the above equation we get,x=14.7m

A ball is dropped from a height of $19.6 m$ . The distance covered by it in the last second is

The correct option is C $14.7 m$
Let x be the distance covered in the last second. Now using $v = u + g t$ , we get
$\sqrt{2 g h} = \sqrt{2 g (h - x)} + g \sqrt{2 g (19.6)} = \sqrt{2 g (19.6 - x)} + g$
Solving the above equation we get,
$x = 14.7$ m