Provide an appropriate response.Derive the equations x = x0 + (1 - e-kt)cos ?, y = y0 + (1 - e-kt)sin ? + (1 - kt - e-kt)by solving the following initial value problem for a vector r in the plane. Differential equation: = -gj - kv = -gj - k Initial

Question

conditions: r(0) = x0i + y0j (0) =  v0 = (v0cos ?)i + (v0sin ?)jThe drag coefficient k is a positive constant representing resistance due to air density, v0 and ? are the projectile's initial speed and launch angle, and g is the acceleration of gravity.



What will be an ideal response?

Answer 1

Problem is separable.
x-coordinate:
  =  = -kv_x or  = -k dt. Integrate to get ln = -kt or
 v_x = v_x,0e^-kt
 Next, x = dt = dt = - e^-kt + C
 At t = 0, x(0) = x₀ = -  + C or C = x₀ + 
 Thus, x = x₀ + (1 - e^-kt). Also, v_x,0 = v₀cos ?, so
 x = x₀ + (1 - e^-kt)cos ?.
 
y-coordinate:
 =  = - g - kv_y. Let w = , then  = . Make substitutions to get:
 = - kw or   = -k dt. Integrate to get: ln = -kt or w = w₀e^-kt.
Replace w:
   = e^-kt. Solve for v_y to get v_y = (g + kv_y,0)e^-kt - .
Finally, y = dt =  dt = - (g + kv_y,0)e^-kt - t + C.
At t = 0, y = y₀ = - (g + kv_y,0) + C or C = y₀ + (g + kv_y,0). Thus,
y = y₀ - (g + kv_y,0)e^-kt - t + (g + kv_y,0) 
 = y₀ + (g + kv_y,0)(1 - e^-kt) - t
Also, v_y,0 = v₀sin ?. Substituting and rearranging:

y = y₀ + (1 - e^-kt)sin ? + (1 - kt - e^-kt)