Question

Two particles A and B are thrown vertically upward
simultaneously, with velocity 5 m/s and 15 m/s
respectively from the same position. The separation
between them after 1 s is (g = 10 m/s2)
(1) 7.5 m
(2) 5 m
(3) 10 m
(4) Zero​

Answer 1

Answer:

(2)s=5m/s

Explanation:

taking concept of relative velocity

net speed is 10

apply second lawof motion

s=ut+1/2at^2

=10-1/2*10*1=5

so 5is your answer

Answer 2

(3) 10 m

$\Delta h=10\ m$ is the separation between the particles A & B after 1 second.

Explanation:

Given:

• initial velocity of A, $u_a=5\ m.s^{-1}$

• initial velocity of B, $u_b=15\ m.s^{-1}$

• time of after which the observation is made, $t=1\ s$

Using the equation of motion:

$v=u-gt$

Velocity after 1 second for object A:

$v_a=5-10\times 1$

$v_a=-5\ m.s^{-1}$

Velocity after 1 second for object B:

$v_b=15-10\times 1$

$v_b=5\ m.s^{-1}$

Now their corresponding distances are given by the below equation of motion:

$v^2=u^2-2g.h$

For A:

$(-5)^2=5^2-2\times 10\times h_a$

$h_a=0\ m$ i.e. the body will be on the ground after 1 second.

For B:

$(5)^2=15^2-2\times 10\times h_b$

$h_b=10\ m$

Now the separation between the two objects will be:

$\Delta h=h_b-h_a$

$\Delta h=10-0$

$\Delta h=10\ m$

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TOPIC: equation of motion

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