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Standard XII

Physics

1st Equation of Motion

A body projec...

Question

A body projected vertically up travels a height $h$ in the $n^{t h}$ second. The distance travelled by it in the next two seconds is

h + 2 g

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2 h + g

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2 h + 2 g

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2 h + 3 g

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Solution

The correct option is D $2 h + 3 g$
$S_{n} = (u (n) - \frac{1}{2} g n^{2}) - (u (n - 1) - \frac{1}{2} g (n - 1)^{2}) = h$
$h = u - \frac{1}{2} g (2 n - 1)$ .....(1)
Distance in next $2$ sec,
$S_{n + 2} = (u (n + 2) - \frac{1}{2} g (n + 2)^{2}) - (u (n) - \frac{1}{2} g (n)^{2})$
$S_{n + 2} = 2 u - \frac{1}{2} g (4 n + 4) - (2)$
$S_{n + 2} = 2 (u - \frac{1}{2} g (2 n - 1)) + 3 g$ .....(3)
$= 2 h + 3 g$ (using 1 in 3)

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