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JEE Main Physics Class 11 Motion in a Straight Line and Plane Part 2 Questions
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© examsnet.com
Question : 94
Total: 100
The trajectory of a projectile in a vertical plane is
y
=
α
x
−
β
x
2
, where
α
and
β
are constants and
x
and
y
are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection
θ
and the maximum height attained
H
are respectively given by
[26 Feb 2021 Shift 2]
tan
−
1
α
,
α
2
4
β
tan
−
1
β
,
α
2
2
β
tan
−
1
α
,
4
α
2
β
tan
−
1
(
β
α
)
,
α
2
β
Validate
Solution:
Given, equation of trajectory of projectile,
y
=
α
x
−
β
x
2
. . . (i)
The standard equation of trajectory of projectile,
y
=
x
tan
θ
−
g
x
2
2
u
2
cos
2
θ
. . . (ii)
Now, from Eqs. (i) and (ii), we get
α
=
tan
θ
and
β
=
g
2
u
2
cos
2
θ
⇒
α
2
=
sin
2
θ
cos
2
θ
∴
β
α
2
=
g
2
u
2
cos
2
θ
×
sin
2
θ
cos
2
θ
=
g
2
u
2
sin
2
θ
×
2
2
4
β
α
2
=
2
g
u
2
sin
2
θ
=
1
H
⇒
H
=
α
2
∕
4
β
θ
=
tan
−
1
α
and
H
=
α
2
∕
4
β
© examsnet.com
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