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A particle is dropped from point A at a certain height from ground. It falls freely under gravity and passes through three points B, C and D with BC=CD. The time taken by the particle to move from B to C is 2 seconds and from C to D is 1 second. The time taken to move from A to B is [Take $g = 10 m / s^{2}$ ]

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0.5s

1.5s

0.75s

0.25 s

answer is B.

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Detailed Solution

Let $h_{B}, h_{C}, h_{D}$ be the distances of points B,C and D from the top of the tower.

From $h = u t + \frac{1}{2} g t^{2 `}$

$h_{B} = \frac{1}{2} g {t_{1}}^{2}$ .........(1)

$h_{C} = \frac{1}{2} g {t_{2}}^{2}$ ..........(2)

$h_{D} = \frac{1}{2} g {t_{3}}^{2}$ ...........(3)

Given BC = CD

$\Rightarrow h_{D} - h_{C} = h_{C} - h_{B}$

$\Rightarrow {t_{3}}^{2} - {t_{2}}^{2} = {t_{2}}^{2} - {t_{1}}^{2}$

Also, given $t_{2} - t_{1} = 2 s e c s$ and $t_{3} - t_{2} = 1 s e c$

From solving the above equations,

we get $2 t_{1} = t_{3} - t_{2} = 1$

$t_{1} = 0.5 s$

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