Projectile launch . . . . . .?

Treating the point of projection as origin, horizontal line through it as x-axis and the vertical line as y-axis, the cartesian equation of the projectile will be

y = x tanθ - 4.9 * (x^2/u^2cos^2 θ) ... ( 1 )

where u = initial velocity and θ is the angle of projection.

=> y = x tan45° - 4.9 * (x^2/500^2cos^2 45°)

=> y = x - (0.0000392) x^2

=> dy/dx = 1 - (0.0000784) x

Distance travelled by it

= ∫(0 to 5000) ds = ∫(0 to x) √[1 + (dy/dx)^2] dx

=> 5000 = ∫(0 to x) √[1 + (1 - (0.0000784x)^2)] dx

=> 5000 = ∫(0 to x) √[2 - 0.0001568 x] dx

=> 5000 = (2/3) * (- 1/0.0001568) * (2 - 0.0001568x)^(3/2) (x = 0 to x)

=> 5000 = - 4251.7 (2 - 0.0001568x)^(3/2) (x = 0 to x)

=> 5000 = - 4251.7 (2 - 0.0001568x)^(3/2) + 4251.7 * 2^(3/2)

=> 5000 = - 4251.7 (2 - 0.0001568x)^(3/2) + 12025.6

=> 7025.6 = 4251.7 (2 - 0.0001568x)^(3/2)

=> (2 - 0.0001568x) = (7025.6/4251.7)^(2/3) = 1.3977

=> 0.0001568x = 2 - 1.3977 = 0.6023

=> x = 3841 m

Plugging this value of x in eqn. ( 1),

y = 3841 - (0.0000392) * (3841)^2

=> y = 3841 - 578 = 3263 m. (Answer)

EDIT:

The length of path = 5000 m is parabolic and √(x^2 + y^2) being chord should be less. However, in my answer above this condition is not satisfied indicating an error in calculations. I checked twice but could not locate the error. However, the method to find the value of y is as explained above.