Which is Faster, Going Up or Coming Down?

1. A ball with mass m is projected vertiaclly upward from Earth’s surface with a positive initial velocity v0 . We assume the forces acting on the ball are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with magnitude p |v(t)|, where p is a positive constant and v(t)is the velocity of the ball at time t. In both the ascent and descent, the total force acting on the ball is -pv -mg. (During the ascent, v(t) is positive and the resistence acts downward; during descent, v(t) is negative and the resistance acts upward.) So, by Newton’s Second Law, the equation of motion is mv' = -pv - mg. Solve this differential equation to show that the velocity is. v(t)= (v0 +(mg/p))e^((-pt/m)) -(mg/p)

2. Show that the height of the ball, until it hits the ground, is y(t)= (v0 +(mg/p)) (m/p) (1-e^(-pt/m))-(mg/p)

3. Let t1 be the time that the ball takes to reach its maximum height. Show that t1= (m/p) ln((mg+pv0)/mg)) Find this time for a ball with mass 1 kg and initial velocity 20 m/s. Assume the air resistance is 1/10 of the speed. In general, it’s not easy to ?nd t2 because it’s impossible to solve the equation y(t) = 0 explicitly. We can, however, use an indirect method to determine whether ascent or descent is faster; we determine whether y(2t1 ) is positive or negative. Show that y(2t1 ) = ((m^2g)/(p^2)) (x-(1/x)-2lnx) where x = e^(pt1 /m). The show that x > 1 and the function f (x) = x - (1/x) - 2 ln x is increasing for x > 1. Use this result to decide whether y(2t1 ) is positive or negative. What can you conclude? Is the ascent or descent faster?

6. What is the answer to the last question in (5) if p=0? justify your answer

 7. Is your answer in (6) consistent with the formula for y(2t1) in (5) and why?