Answer:
Explanation:
Given
Speed of block at bottom is [tex]v=3.8\ m/s[/tex]
Distance traveled [tex]s=6\ m[/tex]
initial velocity is zero
using equation of motion
[tex]v^2-u^2=2as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](3.8)^2-0=2\times a\times 6[/tex]
[tex]a=1.203\ m/s^2[/tex]
when it is 3 m from top then
[tex]v^2-u^2=2as[/tex]
[tex]v^2-0=2\times 1.203\times 3[/tex]
[tex]v=2.68\ m/s[/tex]
