Calculating Angle to Shoot Sword Over Wall for UT - 65 Characters

[Superman123]

Well, if one speed is 5.85 m/s (did the calculation again with 11.6) and the other one is 7.54 then launch speed is

√(5.58^2+7.54^2)= 9.3 m/s

The speed available from the bow is
179=(1.28 kg *v^2)/2
v=16.703

16.703 > 9.3 , meaning that it doesn't exceed the speed available from the bow and that I can use the given speed.

 

[haruspex]

Well, if one speed is 5.85 m/s (did the calculation again with 11.6) and the other one is 7.54 then launch speed is

√(5.58^2+7.54^2)= 9.3 m/s

The speed available from the bow is
179=(1.28 kg *v^2)/2
v=16.703

16.703 > 9.3 , meaning that it doesn't exceed the speed available from the bow and that I can use the given speed.

Looks right.

 

[Superman123]

Thank you very much, I appreciate your help !
Just a very tiny question, did you use suvat-equations to construct this function for the optimal angle (π-atan(d/h))/2?

 

[haruspex]

Thank you very much, I appreciate your help !
Just a very tiny question, did you use suvat-equations to construct this function for the optimal angle (π-atan(d/h))/2?

Yes.