Which angle is this one?

Here's another numerical solution since the one below is pretty conventional.

I have coded in arbitrary values of initial velocity, the value of $g$ , and a value of $\theta$ which is modified.

What I have done is modify the values of theta so that the values of the range of projectile and maximum height approach each other.

I have put some logic at the end of my code to assist with the trial and error and to guide you on whether to increase or decrease the value of $\theta$ .

Try the code yourself if you have MATLAB. In fact, I got the value below (which is extremely accurate) just by trial-and-erroring for 1 minute.

Below are python and matlab codes.

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time = 0;
deltaT = 10^-4; %timestep/accuracy

initialVelocity = 10; %arbitrary initial velocity value

g = 9.81; %gravitational acceleration

theta = 1.3258325; %value of angle in radians (found by trial and error)


%initial values of position
x = 0;
y = 0;

%initial velocities
xDot = initialVelocity*cos(theta);
yDot = initialVelocity*sin(theta);



while yDot >= 0 %simulate till maximum height

    %Explicit Euler ODE solving
    x = x + xDot*deltaT;

    yDotDot = -g;
    yDot = yDot + yDotDot*deltaT;
    y = y + yDot*deltaT;

    time = time + deltaT;

end


maximumHeight = y;  %record maximum height



while y >= 0 %simulate till projectile lands

    %Explicit Euler
    x = x + xDot*deltaT;

    yDotDot = -g;
    yDot = yDot + yDotDot*deltaT;
    y = y + yDot*deltaT;

    time = time + deltaT;

end

range = x; %record range



%Guide for trial and error. 
%Run the code enough times. 
%You'll get an insanely accurate value of theta.    
if range > maximumHeight
        disp("Make theta bigger");
elseif range < maximumHeight
        disp("Make theta smaller");
end


disp("range: ");
disp(range)
disp("height: ");
disp(maximumHeight);

*Here's a python version: *

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import math

time = 0
deltaT = 10**-4 #timestep/accuracy

initialVelocity = 10 #arbitrary initial velocity value

g = 9.81 #gravitational acceleration

theta = 1.3258325 #value of angle in radians (found by trial and error)


#initial values of position
x = 0
y = 0

#initial velocities
xDot = initialVelocity*math.cos(theta)
yDot = initialVelocity*math.sin(theta)#



while yDot >= 0: #simulate till maximum height

    #Explicit Euler ODE solving
    x += xDot*deltaT

    yDotDot = -g
    yDot += yDotDot*deltaT
    y += yDot*deltaT

    time += deltaT



maximumHeight = y  #record maximum height



while y >= 0: #simulate till projectile lands

    #Explicit Euler
    x += xDot*deltaT

    yDotDot = -g
    yDot += yDotDot*deltaT
    y += yDot*deltaT

    time += deltaT



range = x #record range



#Guide for trial and error. 
#Run the code enough times. 
#You'll get an insanely accurate value of theta.    
if range > maximumHeight:
  print("Make theta bigger")
elif range < maximumHeight:
  print("Make theta smaller")

print("range: ")
print(range)
print("height: ")
print(maximumHeight)

$R =\dfrac{u^2 \sin 2 \theta}{g}=\dfrac{2u^2 \sin \theta \cos \theta}{g}$ $H =\dfrac{u^2 \sin^2 \theta}{2g}$

Given,

$R=H$ $\dfrac{2u^2 \sin \theta \cos \theta}{g}=\dfrac{u^2 \sin^2 \theta}{2g}$ $\dfrac{2 \cancel{u^2} \cancel{\sin \theta} \cos \theta}{\cancel{g}}=\dfrac{\cancel{u^2} \cancel{\sin \theta} \sin \theta }{2\cancel{g}}$ $\implies \tan \theta = 4$ $\implies \theta \approx \boxed{75.96}$