It won't hit the wall, I can't throw that hard. Beside all that, I am not a mathematician.

# A Ball Is Thrown With A Velocity U= 40 M/s At An Angle 60 Degrees With The Horizontal Towards A Wall Slanting At An Angle 45 Degrees With The Horizontal. At What Point Will The Ball Hit The Wall?

The vertical height of the ball at time t will be

h

The horizontal displacement of the ball at time t will be

d = 40Cos[60°]t

This horizontal displacement is also the difference between the height of the wall and 20 m.

H

The ball will hit the wall when the ball height h

-16t^2 + 40Sin[60°]t = 20 - 40Cos[60°]t

0 = 20 - t(20+20√3) + 16t^2

The solution to this can be found using the quadratic formula. It will be something like

t = (5(1+√3)-√(20+50√3))/8 ≈ 0.416925

##### The (x, y) coordinates of the point of contact are (d, h

h

_{b}= -16t^2 + 40Sin[60°]tThe horizontal displacement of the ball at time t will be

d = 40Cos[60°]t

This horizontal displacement is also the difference between the height of the wall and 20 m.

H

_{w}= 20 - 40Cos[60°]tThe ball will hit the wall when the ball height h

_{b}is equal to the wall height h_{w}.-16t^2 + 40Sin[60°]t = 20 - 40Cos[60°]t

0 = 20 - t(20+20√3) + 16t^2

The solution to this can be found using the quadratic formula. It will be something like

t = (5(1+√3)-√(20+50√3))/8 ≈ 0.416925

##### The (x, y) coordinates of the point of contact are (d, h_{w}) = (8.3385, 11.6615) meters.

At 75 % angle of depression...

To determine at what point the ball hits the wall you would need the distance to travel of the ball and also need the calculation of the deviation of the flight of the ball and size, shape and weight of the ball.