Question

A paratroops weighing 80kg jumps with zero velocity from an aeroplane at a height of 3000m the air resiatamce encountered by the paratrooper is R(t) =15 v (t2) n where v (t) is the velocity of the paratrooper at time t calcualte the time requrired by the paratrooper to land and tye velocityy at landing

Answer 1

The formula for air resistance is not clear and ambiguous. So I take two cases:

m = 80 kg
Initial Velocity u_y = 0      u_x = 0           h = 3000 m
v(t) = velocity at time t = dy/dt
y = displacement
g = 10 m/s²

The equation of motion: 
Net force on the paratrooper downwards:  m a = m g - R(t)

CASE 1:  
    Air resistance : R(t) = 15 [v(t)]²
    80 dv/dt = m g - 15 v² = 80 g - 15 v²
    dv/dt +  (3/16) v² - 10 = 0
    
If we use the Riccoti differential equation method to solve this, we get 
         v(t) = √(160/3) m/s = constant
         a(t) = 0   
         y(t) = √(160/3) * t

Hence, time required  to land = s/v = 3000 / √(160/3)   sec
     t = 75 √30  sec = 410.79 sec
     v = 7.3  m/sec
========================

CASE 2:  

    R(t) = 15 v(t)
=> Equation of motion :    dv/dt = 10 - (3/16) v
      We find the solution using differential equations method.
      Solution =   v(t) = (160/3) [1 - e^{-(3/16)t } ]  m/s

$v(t)=\frac{160}{3}[ 1-e^{-\frac{3}{16}t} ]\ m/s$

      y(t) = Displacement at t from initial position     (by integration)
             = (160t /3) + (2560/9) e^{-3t/16} - 2560/9    meters

$y(t)=\frac{160}{3}t+\frac{2560}{3}[e^{-\frac{3}{16}t}-1] \ meters$
   
   Solving for t,  with given y = 3000 m,  t ≈ 61.6 sec

   Hence,  v (61.6sec) ≈ 160/3 =  53.33 m/sec,  as paratrooper reaches the ground.