Question

A particle is drop of certain side gravity=0 after 6 another particles is thrown if both particles will meet after 18 second . what value of Velocity​

Answer 1

Answer:

The distance traveled by the first ball in 18s is h=

2

1

gt

2

=

2

1

×10×(18)

2

=1620m

Now to meet second ball has to same distance in(18−6)=12s

So for second ball,

h=vt+

2

1

gt

2

1620=v×12+

2

1

×10×(12)

2

v=

12

1620−720

=75m/s

Explanation:

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