prove that the horizontal range of a projectile is same for two angles of projection
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A Projectile has Same range for 2 diff. angles R1=R2
THIS IMPLIES THAT
Angle of projectile 1 +Angle of Projectile 2 =90(Simple rule Can be proved if required)
Let a + b =90
a=90-b
sin a =cos b
T1 = 2 u sin a/g ;; T1 =2 u cos b/g ; ;cos b=T1 .g /2u
T2= 2 u sin b/g ;; sin b=T2.g/2u
R2 = u*u*sin2b/g
=u*u*2sin b cos b/g
SUBSTITUTE THE VALUES
=u*u*2*T2.g.T1.g/g.2u.2u
=T2*T1*g/2
(WE HAVE ASSUMED THAT THE PROJECTILES HAVE SAME INITIAL VELOCITY
AND ARE PROJECTED FROM SAME POINTS SIMULTANEOUSLY)
THIS IMPLIES THAT
Angle of projectile 1 +Angle of Projectile 2 =90(Simple rule Can be proved if required)
Let a + b =90
a=90-b
sin a =cos b
T1 = 2 u sin a/g ;; T1 =2 u cos b/g ; ;cos b=T1 .g /2u
T2= 2 u sin b/g ;; sin b=T2.g/2u
R2 = u*u*sin2b/g
=u*u*2sin b cos b/g
SUBSTITUTE THE VALUES
=u*u*2*T2.g.T1.g/g.2u.2u
=T2*T1*g/2
(WE HAVE ASSUMED THAT THE PROJECTILES HAVE SAME INITIAL VELOCITY
AND ARE PROJECTED FROM SAME POINTS SIMULTANEOUSLY)
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