# 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?

It won't hit the wall, I can't throw that hard. Beside all that, I am not a mathematician.
thanked the writer.
Vin Ravun commented
Lolz. But it will hit the wall i'm sure.
Hello Boss commented
I try to be entertaining as much as possible. But, watch out when I'm off my Meds, then I'm a Holy terror.
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
##### The (x, y) coordinates of the point of contact are (d, hw) = (8.3385, 11.6615) meters.
thanked the writer.
Vin Ravun commented
Thank you,but i didn't understand why you wrote -16t^2.Could you please explain?
Oddman commented
An object experiencing acceleration "a" moves distance (1/2)*a*t^2 in time "t". Acceleration due to gravity is approximated in these problems by 32 ft/sec/sec in the downward direction. Half that is -16 ft/s^2. Of course, now that I write that, I see that I should have used the metric version of the gravitational constant. The term should be -4.9t^2. The method is right, but the numbers need correction. Sorry about that.
Oddman commented
Revised impact point coordinates are (7.578, 12.422) meters from the origin. Impact is in about 0.3789 seconds. 