A body falling freely under gravity passes two points 30 m apart in 1 s. From what point above the upper point it began to fall? (Take g = 9.8 m .)

Question

A body falling freely under gravity passes two points 30 m apart in 1 s. From what point above the upper point it began to fall? (Take g = 9.8 m $s^{- 2}$ .)

Smart Achievers · Accepted Answer

(1)

Suppose the body passes the upper point at $t$ second and lower point at $(t + 1)$ s, then

$S_{2} - S_{1} = \frac{1}{2} g {(t + 1)}^{2} - \frac{1}{2} g t^{2} = \frac{1}{2} g (2 t + 1)$

or $30 m = \frac{1}{2} \times 9.8 (2 t + 1)$ $∴$ $t = 2.56 s$

$S_{1} = \frac{1}{2} g t^{2} = \frac{1}{2} \times 9.8 \times {(2.56)}^{2} = 32.1 m$

Solution

Q.
A body falling freely under gravity passes two points 30 m apart in 1 s. From what point above the upper point it began to fall (Take g = 9.8 m $s^{- 2}$ .)

1 32.1 m

2 16.0 m

3 8.6 m

4 4.0 m

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