As shown in figure, find that at what distance body A will attain a velocity of 3 m/s when starts from rest.
g=9.81 (m/s2)g=9.81\ (m/s^2)g=9.81 (m/s2)
So, we have
s=v22a=322⋅9.81=0.46 (m)s=\frac{v^2}{2a}=\frac{3^2}{2\cdot9.81}=0.46\ (m)s=2av2=2⋅9.8132=0.46 (m) . Answer
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments