At the high velocity, drag force is proportional to the squared velocity of a particle as kv^2. Find its acceleration when a falling speed becomes half its terminal velocity.
Gives
Drag force
FD="\\frac{1}{2}C\\rho Av^2"
According to question
"F_D=kv^2"
Newton's 2ndlaw
F=Ma
Both force are equal
"Ma=kv^2"
"a=\\frac{kv^2}{M}"
When
"v'=\\frac{v}{2}"
"a'=?"
"a'=\\frac{k(\\frac{v}{2})^2}{M}=\\frac{kv^2}{4M}=\\frac{a}{4}"
Comments
Leave a comment