Show that there are two angles of projection for which the horizontal range is same for
a projectile.
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Explanation:
ANSWER
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
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