Question

2. An object of 5 kg is released from rest 1000 meters above the ground level and allowed to fall under the influence of gravity. Assuming that the force due to air resistance is proportional to the velocity of the object with proportionality constant k = 50 kg/sec determine the formula for the velocity of the object 3. A rocket having an initial mass mo kg is launched vertically from the surface of the Earth. The rocket expels gas at a constant rate of α kg/sec and at a constant velocity of m/sec relative to the rocket. Assume that the gravitational field is a constant g kg/see? Since the mass is not constant, Newton's second law, which states that force is equal to the time rate of change of the momentum, leads to the equation: where u = dr/dt is the velocity of the rocket, z is its height above the surface of the Earth. and mo at is the mass of the rocket at t seconds after launch. If the initial velocity is zero, solve the above equation to determine the velocity of the rocket for 0St<mo/a .A 100 kilogram sailboat is floating motionless in the water. Suddenly, a wind with a constant force of 50 newtons begins to push the boat forward. The force of resistance of the water is proportional to the boats velocity, with a proportionality constant of k = 25 Let v(t) be the velocity of the sailboat after t seconds. a) Write an appropriate differential equation that determines v(t) and solve it b) As time goes on and the force of resistance of the water increases, the velocity of the boat approaches an upper l. Find this upper limit

Answer 1

.

Force equation ma = F_f - F_g since the air resistance (Ff) is proportional to the velocity of the object, it is a laminar flow . Thus Ff will equal kv . with k being the proportionality constant . since v is negative, Ff will be negative as will 5a = -50v - (5) (10) v = -a/10 - 1