Well, then i should suppose that the boy throw the stone with 45 degrees angle... Let's see.

First, to help us in calculations, we convert 90km/h to m/s. This is:

90*1000/3600 = 25m/sec

We call X the horizontal travel, Y the vertical (and what we are looking for...) and V the velocity, Vi the initial and Vf the final, a the acceleration, t the travel time.

X=V

_{IX}t+0.5 x a

_{X} x t

^{2} (1)

V

_{fX}=V

_{iX} + a

_{X} x t (2)

V

^{2}_{fX} = V

^{2}_{iX} + 2a

_{X} x X (3)

these are the formulas for our calculations for the horizontal motion. And we can say that a

_{X}=0, as the stone gets no more acceleration. We rewrite the above as:

(1)=> X=V

_{IX}t (4)

(2)=> V

_{fX} = V

_{iX} (5)

So comes for the Y motion:

Y=V

_{iY} t = 0.5 x a

_{Y} t

^{2} (6)

V

_{fY}=V

_{iY} + a

_{Y} x t (7)

V

^{2}_{fY} = V

^{2}_{iY} + 2a

_{Y} x Y (

When the stone goes to higher point, the vertical velocity V

_{FY} will be 0. so:

(7)=> 0 = V

_{iY} + a

_{Y} x t

and also, the time that the stone will be at the higher point, is half of the total travel time. So, the above equation for the travel from slingshot to max height will result to:

0 = 25ms

^{-1} Sin (45) - (9.8ms

^{-1} x t)/2 => ........ => t=3.607sec.

This will be half the travel time, when the stone is up the sky

Now, we use the equation (5):

(4)=>X=V

_{IX}t => X=25 x Cos (45) x t = ..... => X=63,763 meters.

This is the distance that the boy should stand away from the tree, because at this distance the stone will be at the highest point.

Now with (

(

=> V

^{2}_{fY} = V

^{2}_{iY} + 2a

_{Y} x Y =>

0 = (25 x Sin(45))

^{2} - 2 x 9.8 x Y => .... (solve for Y) ... => Y = 15.94 meters........

If the boy stands 63,763 meters away from the tree and throw the stone with an angle of 45 degrees with 90km/h initial speed, then the stone will go over this tree...

I must admit, i had many many years to deal with projectiles...