Question

ball is dropped from a high rise platform at
= 0 starting from rest. After 6 seconds another
all is thrown downwards from the same platform
with a speed v. The two balls meet at' t = 18 s.
What is the value of v? (Take g = 10 m/s2)
1) 60 m/s
2) 75 m/s
) 55 m/s
) 40 m/s

Answer 75m/s​

Answer 1

Answer:

75m/s

Explanation:

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