Question

a body is allowed to fall freely rest from a height.Find his speed at (a)t =1s (b)t =2s and (c)t =3s after the release.​

Answer 1

Answer:

Initial speed of the bodies u=0.

Distance covered by the first body in t=3 s, S

1

=ut+

2

1

gt

2

Or S

1

=0+

2

1

×9.8×3

2

=44.1 m

Distance covered by the second body in t

′

=2 s, S

2

=ut

′

+

2

1

g(t

′

)

2

Or S

2

=0+

2

1

×9.8×2

2

=19.6 m

Thus distance between the bodies ΔS=S

1

−S

2

∴ ΔS=44.1−19.6=24.5 m

Solve any question of Motion in a

Answer 2

Answer is in the attachment :)

hope it helps you❤