Lets us visualize a situation in which a projectile is fired at a height 1m above the ground and an angle 53 degree to the horizontal ( I can't draw. How u draw ?).



The coordinates of the any arbitrary point be (x, y ).

then,

x = ( horizontal component of vel) * time = u cos 53 * t ---------------(1)



y = (vertical component of vel ) * time --gt²  / 2 = ( u sin 53) *t --gt²  / 2  -----(2)



Putting the value of t from (1) in (2)



y = ( u sin 53) *( x / u cos 53 ) -  g ( x / u cos 53 )²  / 2

or, y = x tan 53  ( gx² ) / 2 u² cos²  (53)

             Now comes the crux of the problem.

Then

y =  p * 1 m -- 1m = p - 1

x = 110 + y

y =  ( 110 + y) tan 53  [ g (110 + y )²  ] /  2 u²  cos²  (53)

y =  ( 110 + y) tan 53  [ 10 ( 12100 + 220y + y² )  ] / 2 * 35 * 35 * cos²   (53)

now

 p - 1 = 5

or, p = 6