A particle is projected on an inclined plane which is inclined at 30^(0) with the horizontal as shown in figure.Initial speed of the particle is v_(0) and inclined plane is sufficiently large. Match the Column- and Column - II. qquad qquad (v_(v))/(30^(@))(v_(0))/(30^(@)) Column-I a) Range on the inclined plane b) velocity of the particle is parallel to the inclined plane at time. d) Time after which particle strikes the plane is d) For the given velocity maximum range on the inclined plane
Answers
Answer:
Given : u=10 m/s
Initial velocity along x direction u
x
=ucos30
o
m/s
Initial velocity along y direction u
y
=usin30
o
m/s
Also acceleration x and y direction a
x
=−gsin30
o
m/s
2
and a
y
=−gcos30
o
m/s
2
x direction : V
x
=u
x
+a
x
t
time t=
10cos30
2×10sin30
, hence t=
3
2
s
∴ V
x
=ucos30
o
−(gsin30
o
)t
OR V
x
=10×
2
3
−10×0.5×
3
2
=
3
15−10
=
3
5
m/s
Also V
y
=−5 m/s (solved earlier)
∴ Speed of the particle at collision V=
V
x
2
+V
y
2
=
(
3
5
)
2
+(−5)
2
⟹ V=
3
100
=
3
10
m/s
Answer:
Given : u=10 m/s
Initial velocity along x direction u
x
=ucos30
o
m/s
Initial velocity along y direction u
y
=usin30
o
m/s
Also acceleration x and y direction a
x
=−gsin30
o
m/s
2
and a
y
=−gcos30
o
m/s
2
x direction : V
x
=u
x
+a
x
t
time t=
10cos30
2×10sin30
, hence t=
3
2
s
∴ V
x
=ucos30
o
−(gsin30
o
)t
OR V
x
=10×
2
3
−10×0.5×
3
2
=
3
15−10
=
3
5
m/s
Also V
y
=−5 m/s (solved earlier)
∴ Speed of the particle at collision V=
V
x
2
+V
y
2
=
(
3
5
)
2
+(−5)
2
⟹ V=
3
100
=
3
10
m/s
solution