A spherical mass 'm'is dropped vertically off a building starting from rest and 'y=0'.the mass experience a resistive force 'FR=-bv'.let 'r=b/m'.find the position'y(t)'of mass??? pizzzzzzzzzzz help me.........
To find the position y(t) of the mass, we need to solve the differential equation that relates the resistive force FR with the position y. The equation is given by:
m(d^2y/dt^2) = -bv
To solve this equation, we can introduce a variable v = (dy/dt). This allows us to rewrite the equation as:
m(dv/dt) = -bv
To separate the variables, we can rearrange the equation as:
dv/v = (-b/m)dt
Integrating both sides gives us:
∫(1/v) dv = ∫((-b/m)dt)
ln(v) = (-b/m)t + C1
Exponentiating both sides gives us:
v = e^((-b/m)t + C1)
Next, we can solve for y by integrating the expression for v with respect to t:
dy/dt = v
dy = v dt
Integrating both sides gives us:
∫dy = ∫v dt
y = ∫(e^((-b/m)t + C1)) dt
Integrating e^((-b/m)t) dt gives us:
y = (-m/b) * e^((-b/m)t) + C2
After integrating and simplifying, we find that the position y(t) of the mass is given by:
y(t) = (-m/b) * e^((-b/m)t) + C2
where C1 and C2 are constants that can be determined by the initial conditions of the problem.