Please help ............Thank you

Question

The terminal velocity of a 3×10^−5 kg raindrop is about 9 m/s . Assume a drag force FD=−bv.

a) determine the value of the constant b (one sig figure)

b)Determine the time required for such a drop, starting from rest, to reach 63% of terminal velocity.(one sig figure)

Answer 1

Taking our known values we know that v_t (v_terminal) = 9 m/s, m (mass) = 3*10^−5 kg. Since we also know we are dealing with a falling object affected by gravity it can be helpful to note the value for gravitational acceleration (g) is 9.8 m/s^2. Now let us consider the free body diagram for this system where the downward y-direction will be positive for us. Then we know F_d (the drag force) is in the negative direction and that the weight of the object (m*g) is pulling the droplet in the positive downward direction.

a) Newton’s second law in the vertical direction gives the differential equation

m*g−b*v= m*a = m*(dv/dt). equation 1

Now if we take this equation to its limit, where we accelerate to terminal velocity, there will be no acceleration (dv/dt = 0) because our velocity remains constant. This allows us to write

m*g - b*v_t = 0 OR m*g = b*v_t.

From here you may solve for the unknown variable b and plug in your known values.

b) To determine the time required for such a drop, starting from rest (v_i = 0, initial velocity), to reach 63% of terminal velocity we can integrate our previous equation, equation 1, but first we must manipulate it into the form of

dt = dv/(m*g-b*v).

Here we can integrate the left-hand-side over dt from 0 to some time t that we want to find. Then we integrate the right-hand-side over dv from 0 to the velocity of interest, or v_f or the final velocity. In this case we want to determine the time it will take for the droplet to accelerate from 0 to 63% of v_t so this will be the final velocity we plug in. The final equation before plugging in numbers should look like this

t = −(m/b)*[ln(g−(b/m)v)−ln(g)].