Question

Consider an object with mass m that is thrown with an initial speed 0 at an angle θ with respect to the ground. The gravitational acceleration is g and it can be assumed to be constant. Derive the differential equations, and initial conditions, that govern the horizontal and vertical motion of the object assuming
(a) no air resistance and
(b) air resistance is proportional to the speed of the object.

Answer 1

Answer:

a) dx = v₀o cos θ dt ,   dy =  (v_{oy} - g t) dt

b)  dx / dt = v₀ₓ  e^{- (\gamma/ m)  t} ,   dv_y/dt - γ/m v_y + g =0

Explanation:

For this exercise we will use Newton's second law on each axis

       F = m a

       F = m d²x / dt²

let's use trigonometry for the components of the initial velocity

        cos θ = v₀ₓ / v₀

        sin θ = v_{oy} / v₀

        v₀ₓ = vo cos θ

        v_{oy} = v₀ sin θ

now we will work on each component

a) in this case there is no air resistance

X axis

     Fₓ = 0

       0 = m d²x / dt²

       0 = m dv / dt

we integrate once

        m dv = 0

        m (vf- vo) = 0

       v_f = v₀

we use the definition of velocity

       v = dx / dt

       vₓ = dx / dt

       dx = v₀ₓ dt

       dx = v₀o cos θ dt

we integrate once

        x_f-x₀ = vₓ cos θ (t-0)

y axis

in this case the force is the weight of the body

        F = W = - mg

        F = m a

         -m g = ma

        a = -g

we use the relationship between acceleration and velocity

         a = dv / dt

          dv / dt = -g

         dv = -g dt

we integrate

         (v_y-v_{oy}) = -g (t-0)

          v_y = v_{oy} - g t

now we love the relationship

         v = dy / dt

         dy / dt = v_{oy} - g t

        dy =  (v_{oy} - g t) dt

we integrate

       y- y_o = v_{oy} - ½ g t²

      y = y_o + v₀ sin tea t - ½ g t²

B) in this case there is air resistance, we will assume that in the two axes

x-axis) resistance always opposes movement

            fr = - g v

           

             ∑ F = m aₓ

           - γ vx = m dvₓ / dt

              dvₓ / vₓ = - γ/m  dt

              ln vₓ / ln v₀ₓ = - γ/m   (t-0)

              vₓ = v₀ₓ $e^{- (\gamma/ m) t}$

let's use the relationship between speed and distance

              dx / dt = v₀ₓ  e^{- (\gamma/ m)  t}

              dx = v₀ₓ e^{- (\gamma/ m)  t}  dt  

               

Y axis

       ∑ F = m a_y

        fr- W = m a_y

        γ v_y - m g = m a_y

        a_y =   γ/m  v_y  - g

        d v_y / dt = (γ /m  v_y - g)

        dv_y/dt - γ/m v_y + g =0

       

in this equation differences for the initial time the velocities vo = vo sin θ