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Two stones A & B are thrown simultaneously as shown from a very high tower. Neglect air friction
List- I List – II
A)
P)
Distance between the two balls after two seconds is
$8 \sqrt{5} m .$
B)
Q)
Distance between two balls after 2 seconds is
40 m.
C)
R) Magnitude of relative velocity of B with respect to A is $5 \sqrt{5} m / s .$
D)
S)
Magnitude of relative velocity of B w.r.t A is
$5 \sqrt{2} m / s .$
T)
Magnitude of relative velocity of B with
respect to A is $10 \sqrt{5} m / s .$
Then the correct match of the lists is

see full answer

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A – S, B – P, C – R, D – Q

A – Q, B – R, C – P, D – Q

A – S, B – R, C – T, D – P

answer is B.

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Detailed Solution

If the relative velocity is $v = v_{x} \hat{i} + v_{y} \hat{j}$

Distance between the 2 balls after 2 seconds is

$X = v_{x} t = 2 v_{x}$

$Y = v_{y} t = 2 v_{y}$

$S = \sqrt{X^{2} + Y^{2}} = 2 \sqrt{{v_{x}}^{2} + {v_{y}}^{2}}$

Magnitude of relative velocity is

$\sqrt{{v_{x}}^{2} + {v_{y}}^{2}}$

For the given cases,

(i) Relative velocity is $- 20 \hat{j}$

$S = 40 m$ , $|V| = 20 m / s$

(ii) Relative velocity is $10 \hat{i} - 10 \hat{j}$

$S = 40 \sqrt{2} m$ , $|V| = 20 \sqrt{2} m / s$

(iii) Relative velocity is $8 \hat{i} - 4 \hat{j}$

$S = 16 \sqrt{5} m$ , $|V| = 4 \sqrt{5} m / s$

(iv) Relative velocity is $16 \hat{i} - 12 \hat{j}$

$S = 40 m$ , $|V| = 20 m / s$

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