Question

Show that there are two angles of projrction for same horizantal range.

Answer 1

Let the velocit yo f projection=U 
and angle of projection=θ 
The vertical component of velocity =Usinθ 
Horizontal component of velocirty= Ucosθ 
Equationof vertical motion is 
H=Usinθt−1/2 gt² 
Time of flight, t =2Usinθ/g 
Equation of horizontal motion is 
S = Ucosθ t 
Hence S =Ucosθ×2Usinθ/g=U²sin2θ/g=U²sin(180−2θ)/...‡ 
Hence there are two valus of teh angleof projection for the same horizontal range 
i.e θ and (180−2θ)/2 
i.e θ and 90−θ

Answer 2

horizontal range is given by u²sin2x/g
sin2x can be same for two angles of the form 45±y
for ex for launch angles 30° & 60° sin2x is √3/2