1. Derive the equations x = [1 - e-kt] (cos a) g y = - +...

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1. Derive the equations x = [1 - e-kt] (cos a) g y = - + -[1 ekt](sin a) + 1/2 (1 kt ekt) - by solving the following initial value problem for a vector r in the plane Differential equation: dr dt = gj kv = gj k - dt r(0)=0 dr dt V = (vo cos )i + (vo sin )j t=0 The drag coefficient k is positive constant representing resistance due to air density, vo and a are the projectile's initial speed and launch angle, and g is the acceleration due to gravity. 1. Derive the equations x = [1 - e-kt] (cos a) g y = - + -[1 ekt](sin a) + 1/2 (1 kt ekt) - by solving the following initial value problem for a vector r in the plane Differential equation: dr dt = gj kv = gj k - dt r(0)=0 dr dt V = (vo cos )i + (vo sin )j t=0 The drag coefficient k is positive constant representing resistance due to air density, vo and a are the projectile's initial speed and launch angle, and g is the acceleration due to gravity.

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