Find the angle with which the projectile was launched?

Questions
,
1

I need to find angle(?) in which projectile was launched but I have no idea which physics formulas should I use. I have: v0=930m/s, R=100m, b=50m and g=9.8m/s^2 I have to find: ?=?

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2

Try this formula

y = x Tan[Theta] - (g x^2 (1 + (Tan^2)[Theta]))/(2 v^2)

substituting your letters would be

b = R Tan[Theta] - (g R^2 (1 + (Tan^2)[Theta]))/(2 v0^2)

and with the figures the angles are

ArcTan[.500709] and ArcTan[1762.8]

or 26.5975 degrees and 89.9674

3

x-dir

 Vx=v0*Cos[theta]
 ?d=R=100(* meters*)
 ?t=t
 Vx=?d/?t
 v0*Cos[theta]=?d/?t=R/t
 t=R/v0*Cos[theta]

y-dir

  Vy=v0*Sin[theta]
   a=-g
   ?d=b=50(* meters*)
   ?t=t
   ?d=v*t+1/2*a t^2
   b=v0*Sin[theta]*t-g/2*t^2

then :

   b=v0*Sin[theta]*(R/v0*Cos[theta])-g/2*(R/v0*Cos[theta])^2

and then:

  v0 = 930; R = 100; b = 50; g = 9.8;
  sol = Theta /. Solve[b == R*Tan[Theta] - g/2*(R/(v0*Cos[Theta]))^2, Theta] /. C[1] -> 0
  (* {-2.67738, -1.57136, 0.464214, 1.57023}*)
  {sol[[3]]/Degree // N // DMSString, sol[[4]]/Degree // N // DMSString}
  (* 26°35'51.075" or 89°58'3.110" *)

Reference.

With high speed bullet and for a very short distance formula is very easy:

R = 100;
b = 50;
ArcTan[b/R]/Degree // N // DMSString

(* 26°33'54.184'' *)