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
hb = -16t^2 + 40Sin[60°]t
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.
Hw = 20 - 40Cos[60°]t
The ball will hit the wall when the ball height hb is equal to the wall height hw.
-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
hb = -16t^2 + 40Sin[60°]t
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.
Hw = 20 - 40Cos[60°]t
The ball will hit the wall when the ball height hb is equal to the wall height hw.
-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, hw) = (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.
