Question

A stone is dropped from the roof of a high building. A second stone is dropped 1s later. how far apart are the stonses when the second one has reached a speed of 40m/s ?​

Answer 1

Answer:

v=u+at or 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

Answer 2

Answer:

Answer

Correct option is

D

40m

v=u+at or 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