A cabin is moved up the inclined plane with constant acceleration g sin x. A particle is projected with same velocity with respect to the cabin in a direction perpendicular to the inclined plane. If maximum height attained by the particle perpendicular to the inclined plane is same as range of particle with respect to the cabin parallel to the plane then calculate values of cot x
Answers
Answer:
Angle of inclination of plane:
θ
=
A
Velocity of cabin after time
t
1
:
g
sin
A
×
t
1
Initial velocity of projectile perpendicular to the plane:
u
y
=
g
sin
A
×
t
1
Initial velocity of projectile along the plane:
u
x
=
0
Acceleration perpendicular to the plane:
a
y
=
g
cos
A
Acceleration along the plane:
a
x
=
g
sin
A
Expression of time of flight is
t
2
=
2
u
y
a
y
Write the expression for the range of the projectile
R
=
u
x
t
2
+
a
x
t
2
2
2
R
=
u
x
(
2
u
y
a
y
)
+
a
x
(
2
u
y
a
y
)
2
2
R
=
2
u
x
u
y
a
y
+
2
a
x
u
y
2
a
y
2
.
.
.
(
i
)
Write the expression for maximum height
y
=
u
y
2
2
a
y
.
.
.
(
i
i
)
Since the range and height are equal, by equating equation (i) and (ii) we get
u
y
2
2
a
y
=
2
u
x
u
y
a
y
+
2
a
x
u
y
2
a
y
2
u
y
=
4
u
x
+
4
a
x
u
y
a
y
Substitute values in the above expression
g
sin
A
×
t
1
=
0
+
4
g
sin
A
×
g
sin
A
×
t
1
g
cos
A
1
=
4
sin
A
cos
A
cot
A
=
4
Answer:
1)Write a program to input any number check
it is positive,negative or neutral.
Display the result