Modeling motion with air resistance (integral calculus)

[numba1stunna]

Please, any help would be appreciated.

Air resistance is a force that acts in the direction opposite to the motion and increases in magnitude as velocity increases, let us assume at least initially that air resistance r is proportional to the velocity: r = pv, where p is a negative constant. suppose a ball of mass m is thrown upward from the ground. The net force f on the ball is F = r – mg (the direction of the force r is downward (negative) when the ball is traveling upward and the direction of r is upward when the ball is traveling upward.)

1. Use the net force equation and the fact that F = ma to write a differential equation for the ball’s velocity.

2. Assume m = .5kg and p = .1. Make a direction field for the differential equation and sketch a solution of the initial value problem v(0) = 50 m/s.

3. Solve the initial value problem algebraically. Hint: be sure to take the constant of integration into account.

4. Find an equation of the height of the ball at time t.

5. When does the ball reach the apex of its trajectory? When does the ball land?

6. Does it take the ball longer to come up or come down?

This is what I've done so far.
m(dv/dt) - pv - mg = 0 for the upward motion

and

m(dv/dt) + pv - mg = 0 for the downward motion

taking upward to be positive (and both p, g < 0).

 

[lippyka]

1. m(dv/dt) + pv - mg = 0 2. Direction Field: The direction field for this equation is the same for both upward and downward motion and is given by the graph below: 3. Solving algebraically: The solution to the initial value problem is given by v(t) = 50 e^(-pt/m) - gt/p. 4. The equation of the height of the ball at time t is given by h(t) = 50t - (1/2)(gt^2)/p + c, where c is the constant of integration. 5. The ball reaches the apex of its trajectory when v(t) = 0, so t = (50m/g)e^(-pt/m). The ball lands when h(t) = 0, so t = [-c + (50/g)e^(-pt/m)]/(1/2)(g/p). 6. It takes the ball longer to come down than it does to come up.