Answer:
a. v(t)= -6.78 + 16.33 b. 16.33 m/s
 + 16.33 b. 16.33 m/s 
Step-by-step explanation:
The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)= =
 = . We now multiply both sides of the equation by the integrating factor.
. We now multiply both sides of the equation by the integrating factor. 
μv' + μkv = μg ⇒  v' + k
v' + k v = g
v = g ⇒ [v
 ⇒ [v ]' = g
]' = g . Integrating, we have
. Integrating, we have 
∫ [v ]' = ∫g
]' = ∫g
     v =
 = 
 + c
 + c
     v(t)=    + c
 + c .
. 
From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have 
9.55 = 9.8 × 15/9 + c = 16.33 + c
 = 16.33 + c
        c = 9.55 -16.33 = -6.78.
So, v(t)=   16.33 - 6.78 . m/s = - 6.78
. m/s = - 6.78 + 16.33 m/s
 + 16.33 m/s
b. Velocity of object at time t = 0.5
At t = 0.5, v = - 6.78 + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s
 + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s