Answer: The power required by the pump to produce a discharge of 9m³/s is 5990joules/secs.
Explanation: The given parameters from the questions are:
Flow rate Q = 9m³/s, Diameter D = 0.98m, acceleration a = 1.0, head loss(Pressure P) is given by the function 10v²/2g.
STEP 1. Find the velocity of water in the pipe from the equation:
Diameter D = (√4.Q/π.v), where v is the velocity, and Q is flow rate
Making v subject of the formula gives:
v = 4Q/π.√D =[ 4 × 9m³/s / 3.142 × (√0.98m)] = 11.69m/s.
STEP 2. Find the pressure from the relationship, P = 10v²/2g, NB. g = a
P = 10 × (11.69m/s)² / 2× 1.0m/s²
P = 683.25N/m² or Pascal.
STEP 3. Find force exerted by the pump;
Recall that Pressure P = Force/Area
But Area A = π.r², where r = D/2
Therefore, A = π.(D/2)²
A = 3.142 × [0.98m/2]² = 0.75m²
Therefore, Force = Pressure × Area
Force F = 683.25N/m² × 0.75m²
F = 512.44N.
STEP 4. Find work done
Work done W by the pump is = Force × distance d moved by the water
W = F . d
Also recall that flow rate Q = Velocity/time.
Q = v/t, we can write t = v/Q.
Time t = 11.69m/s / 9m³/s = 1.298s
Also recall that velocity v = distance d/time t, v = d/t, making d subject of formula gives v × t
Distance d = v × t = 11.69m/s × 1.298s = 15.17m.
Hence,
Work Done W = Force × distance
W = 512.44N × 15.17m = 7775.56Nm or joules.
Lastly, Power P = Work done/ time
P = 7775.56joules/1.298s
P = 5990.4joules/s.
and plug in what we know. For 
, we know it is zero because at the highest location of flight, there is no vertical velocity. For 
, we know it is 
since the vertical component is the sine of the velocity. We need to solve for 
, so it is left as-is. Since the only acceleration is 
, we can substitute -9.81 into it. Solving for the equation yields a solution of 17.55 m for 
.