Trajectory Evaluation

Here's a python solution.

Neeraj, what I have done here in the code is as follows:

First step: Set the condition to when the object hits the ground, when $y >= 0$ .

Second step: Calculate the Kinetic and Potential energy of the object during trajectory by finding magnitude of velocity and dividing by 2, and potential by finding $mgh$ .

Third: Increment the value of $x$ every $10^{-7}$ seconds. This will be done by multiplying the time by the velocity, does adding small distances to the current distance. On a more technical perspective this is called Euler Integration.

4rth: Do the same steps to the $y$ value, by calculating velocity using the acceleration and then numerically integrating to get the displacement.

5th: using the computations, numerically integrate to find the integrals of the Kinetic and Potential Energy.

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import math


#time values and increment
time = 0
deltaT = 10**-7 

#x and y
x = 0
y = 0

#velocities in both directions (arbitrary)
xDot = 10
yDot = 10

#arbitrary g value
g = 10


#Kinetic Energy and potential arbitrary initialisation
integralKinetic = 0
integralPotential = 0

while y >= 0:
    #kinetic and potential computation
    Kinetic = (xDot**2+yDot**2)/2
    Potential = g*y

    #acceleration in y
    yDotDot = -g

    #updating x value
    x += xDot*deltaT

    #numerically integrating yDot to update y value
    yDot += yDotDot*deltaT
    y += yDot*deltaT

    #Kinetic energy increment and Potential increment
    dK = Kinetic*deltaT
    dP = Potential*deltaT

    #adding dKs and dPs to get the resulting integral
    integralKinetic += dK
    integralPotential += dP

    time += deltaT

print(integralKinetic/integralPotential)

Python code based solution:

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import math

# Parameter initialisation:
m     = 1
u     = 1
theta = math.pi/4
g     = 10

# Time step:
dt    = 10**-6

# Variables initialisations:
t     = 0
x     = 0
y     = 0
dx    = u*math.cos(theta)
dy    = u*math.sin(theta)

# Integrals of KE and PE wrt time initialisation:
TS    = 0
VS    = 0

# Loop for numerical integration:
while y >= 0:

    # Acceleration components:
    ddx = 0
    ddy = -g

    # Kinetic and potential energies:
    T   = 0.5*m*(dx**2 + dy**2)
    V   = m*g*y

    # Numerical integration to obtain velocity components:
    dx  = dx + ddx*dt
    dy  = dy + ddy*dt

    # Numerical integration to obtain position coordinates:
    x   = x + dx*dt
    y   = y + dy*dt

    # Numerical integration to obtain Integrals of KE and PE respectively:
    TS  = TS + dt*T
    VS  = VS + dt*V

    t   = t + dt

# Required ratio:
ANSWER = TS/VS
print(ANSWER)
# 2.0000348700607242

From this Problem we know that the generalise value of $\frac{\int_{0}^{t_{f} } T(t) dt} {\int_{0}^{t_{f} } V(t) dt } = \alpha$
comes out to be $\Large \frac{\sin \theta+ \frac{4 \sin^{3} \theta }{3} -2\sin^{3} \theta}{ \frac{2 \sin^{3} \theta }{3}} = \alpha$
simpifying the $\alpha$ leads to $\large \alpha = \frac {3 }{2\sin^{2} \theta}-1$
Now ,lets put $\theta =45$ degrees $\boxed{ \alpha = 2}$

We can show (see this ) that the ratio is

$\eta=\dfrac {3}{2\sin^2 \theta}-1$

Here $\theta=45\degree$

So $\eta=\dfrac {3}{2\times \frac 12}-1=3-1=\boxed 2$ .