Question

Velocity of a stone dropped from the top of the building after 2 seconds is
(Take g = 10 m/s2)

​

Answer 1

Answer:

here ,

initial velocity (u ) = 0m/s

time , t = 2 sec

g = 10m/s²

final velocity , v = ....?

from the equation

v = u + gt

=> v = o + 10* 2

=> v = 20 m/s

so, the. velocity of the body is 20m/s

$..$

I Hope it will help you

if it helps you then Mark as brainliest...

Answer 2

Answer:

v=u+at v=9.8t

t=

10

30

=3.0s is the time at which stone will reach 30 m/s velocity

First stone will reach s=ut+(1/2)at

2

so s

1

=1/2×10×9m

s

1

=45m

(initial velocity = u = 0)

second stone was dropped after 2sec, so time for itt

1

=3.0−2.0s=1.0s

s

2

=1/2×10×1

2

=5.0m

so distance between two stones

D=45.0−5.0=40.400m