# Find the angle with which the projectile was launched?

Posted 8 days ago
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 I need to find angle(φ) 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: φ=?
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Posted 8 days ago
 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
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Posted 8 days ago
 x-dir  Vx=v0*Cos[theta] Δd=R=100(* meters*) Δt=t Vx=Δd/Δt v0*Cos[theta]=Δd/Δt=R/t t=R/v0*Cos[theta] y-dir  Vy=v0*Sin[theta] a=-g Δd=b=50(* meters*) Δt=t Δd=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'' *) 
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