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 $g = 10 m / s^{2}$ ]

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

0.5s

1.5s

0.75s

0.25 s

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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

$t_{1} = 0.5 s$

Watch 3-min video & get full concept clarity