Dear Taparsranjan

first find out the equation of parabola

let highest point is  origin. so equation of parabola : x²=-ay (opening downward)

  point( 50,-5) will lie on parabola

so a will come a=500

 x²=-500y

now differentiate

2x=-500 dy/dx          so dy/dx at x=0  is 0

and 2= -500d²y/dx²   so d²y/dx²  at x=0 is -1/250

radius of curvature at x=0 of parabola 

1/R  = d²y/dx²  /[ 1+(dy/dx)²]^3/2

 R= 1/d²y/dx²

    ⁼ 250

force on bridge at highest point = mg-mv²/R

                                            =1000 * 10  - 1000*(20)²/250

                                            =8.4 kN in the downward  direction

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