Particle-1 is just dropped from a tower. $1 s$ later particle-2 is thrown from the same tower horizontally with velocity $10 m s^{- 1}$ Taking $g = 10 m s^{- 2},$ match the following two columns at $t = 2 s$ .

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Solution

Assuming particle_1 is dropped at $t = 0 s$
therefore at $t = 2 s$
$h_{1} = u_{1} t + \frac{1}{2} g t^{2}$
$h_{1} = \frac{1}{2} \times 10 \times 4$
$h_{1} = 20 m$
$h_{2} = u_{v} (t - 1) + 1 / 2 g (t - 1)^{2}$
$h_{2} = 1 / 2 \times 10 \times 1$
$h_{2} = 5 m$
relative verticle displacement = $h_{1} - h_{2} = 15 m$
Now, horizontal displacement of 1 is zero.
$R_{1} = 0$
$R_{2} = u_{h} (t - 1) = 10 \times 1 = 10 m$
relative horizontal displacement= $R_{2} - R_{1} = 10 m$
relative horizontal velocity $u_{r h} = u_{h} - 0 = 10 m / s$ since for particle 1 horizontal velocity is zero.
Now verticle velocity of 2 at t=2s
$u_{v} = u_{2 v} + g (t - 1) = 10 m / s$
and $u_{1} = u_{1 v} + g t = 20 m / s$
therefore,relative vertical velocity $u_{r v} = u_{1} - u_{v} = 10 m / s$
therefore,
$A \to 1$
$B \to 4$
$C \to 1$
$D \to 1$

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