Question

k III.pdf 1) Up until now we have always ignored air resistance. We should now add . Let us just think of simple 1-dimensional problem, dropping a ball of mass m from a height H but 2 with air resistance. We can model the air resistance as a force proportional to the velocity, fair = bu. The coefficient b is a constant. (For this problem you can use calculus textbooks or wolfram alpha to do the calculus.) What are the units on b? . • Why does it make sense for the force to be dependent on the velocity? The constant b depends on other physical parameters, can you think of what these might be? I only gave the magnitude of the air resistance force. What is the direction? Choose a standard coordinate system with the j direction being vertical and positive up. Write out Newton's second law for this problem. Since this is a 1-dimensional problem you don't have to worry about vector notation. • When we derived our kinematic equations we integrated Newton's second law, essen- tially solving for the velocity and position from a known function for the acceleration. The kinematic equations assumed what abut the acceleration? How is this problem different? . When the air resistance force and gravitational force balance out the acceleration goes to zero. When the acceleration goes to zero there is no longer a change in motion, hence the velocity does not change (right?). When this happens we say the object has reached terminal velocity, vt. Find an expression for the terminal velocity given the parameters of the problem. • Following the procedure similar to the constant acceleration case, find the velocity as a function of time by integrating F = ma. Plot this function. Integrate your velocity v = dy/dt to get y as a function of time. Plot this function. .

Please answer the last 4 bullet points step by step. Again the last 4 bullet points.

Answer 1

Newton's as m du dt for this problem can be written mg- br We assumed that acceleration is constant. But here this is not the case, as acceleration (dy) depends on V there is change in velocity hence dv at = 0; eqh (i) mg bv. = mg at v=vt no o INT ma b m du dt mg-br mdu b/mg dt b du mg عاه dt -v) b mg dv t1 V dt SS bt VT dv v-v Infv. Vp) » VVT v=Vrtce tlnc m " Integrating cont - ce-bt/m - bt/m

at t=0 v=0 o tc & Ca - Vp hence V= p - vp e-btlm Vp (1-e - balm) >V= t plot of V us at V- Since dy at dy dt 4 (1-e-bilm) - but lm) dt → Sdy - Sucre v (t + e- bet m m ] + c bt/mpe ] + C integration confort,

at t= 0, y = 0 my tc'-o c'--my b b => y = ve (attm .bt/m)-mv, e- b b > y = vytt mu ho ut [2 - e-bt/m This is hight fallen from initial position after time it' since intial hence its height at time at H-Y H- Up at tm vt [2- e-btlon a height was h b Н YT y decreases almost linearly as ut dominates e-bt/m for small b, t

Best of luck with the rest of your coursework ?