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Byju's Answer
Standard XII
Mathematics
Perpendicular Distance of a Point from a Line
A particle is...
Question
A particle is projected with speed
u
at angle
θ
with vertical. Find range.
A
R
=
u
2
sin
2
θ
2
g
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B
R
=
u
2
cos
2
θ
2
g
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C
R
=
u
2
sin
2
θ
g
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D
R
=
u
2
cos
2
θ
g
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Solution
The correct option is
C
R
=
u
2
sin
2
θ
g
Time of flight ,
T
=
2
u
c
o
s
θ
g
∴
Horizontal range,
R
=
u
x
T
=
(
u
s
i
n
θ
)
×
2
u
c
o
s
θ
g
=
u
2
(
2
s
i
n
θ
c
o
s
θ
)
g
=
u
2
s
i
n
(
2
θ
)
g
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Similar questions
Q.
A particle is projected with speed
u
at an angle
θ
with the horizontal. Find the radius of curvature at the highest point of its trajectory.
Q.
Assertion :When a body is projected at an angle
45
0
, its maximum height is half of horizontal range. Reason: Horizontal range
=
U
2
s
i
n
2
θ
g
and maximum height
=
U
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s
i
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Q.
A particle is projected from the ground with velocity
u
at an angle
θ
with horizontal. The horizontal range, maximum height and time of flight are
R
,
H
and
T
respectively. They are given by,
R
=
u
2
sin
2
θ
g
,
H
=
u
2
sin
2
θ
2
g
and
T
=
2
u
sin
θ
g
Now keeping
u
as fixed,
θ
is varied from
30
∘
to
60
∘
. Then
Q.
Show that the maximum height and range of a projectile are
u
2
s
i
n
2
θ
2
g
and
u
2
s
i
n
2
θ
g
respectively
Q.
Assertion :When range of a projectile is maximum, its angle of projection may be
45
0
or
135
0
. Reason: Horizontal range
R
=
u
2
sin
2
θ
g
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, when
θ
=
45
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or
135
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