prove that horizontal range is same for two different angle
Answers
Answer:
Motion in A Plane. Show that there are two angles of projection for which horizontal range is the same. Thus horizontal range is the same for angle of projection θ and α = 90°–θ.
Explanation:
Let the two angles be θ
1
and θ
2
and the projectile velocity be u
Horizontal range R=
g
u
2
sin(2θ
1
)
=
g
u
2
sin(2θ
2
)
=
g
u
2
sin(180−2θ
2
)
⟹ 2θ
1
=180
o
−2θ
2
⟹ θ
2
=90−θ
1
Maximum height attained H
1
=
2g
u
2
sin
2
θ
1
Similarly H
2
=
2g
u
2
sin
2
θ
2
=
2g
u
2
sin
2
(90−θ
1
)
=
2g
u
2
cos
2
θ
1
∴ H=H
1
+H
2
=
2g
u
2
(sin
2
θ
1
+cos
2
θ
2
)
=
2g
u
2
Answer:
R₁ = R₂ ( Ranges are equal )
Explanation:
To prove that horizontal range is same for two different angle, let us consider that speed of projection is constant for both the projectile. Also, let the angle of projection for projectile₁ be ∅ and for projectile₂ be 90 - ∅. Additionally, let the two horizontal ranges be R₁ and R₂.
Now,
R₁ = u² sin (2∅)/ g ________{1} (from the formula for horizontal range)
and, R₂ = u² sin 2( 90 -∅)/ g ________{2} (from the formula for horizontal range)
From {2}, we have;
R₂ = u² sin 2( 90 -∅)/ g
R₂ = u² sin (180 - 2∅)/ g
R₂ = u² sin (2∅)/ g ____{3} [Since, sin(180-α)= sinα}
Now comparing {1} and {3} we conclude that;-
R₁ = R₂
Hence, proved!
Hope it helps ;-))