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Byju's Answer
Standard XII
Physics
Motion Under Gravity
A particle is...
Question
A particle is projected with a velocity {\vec v=(3^ i-^ j+2^ k)m/s and a constant acceleration { acting on the particle is \vec a=(-6^ i+2^ j-4^ k)m/s Then the path of projectile is
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Similar questions
Q.
A particle is moving in a circular path. The acceleration and momentum vectors at an instant of time are
→
a
=
2
→
i
+
3
→
j
m
/
s
2
and
→
P
=
6
→
i
−
4
→
j
kgm/s. Then the motion of the particle is:
Q.
A particle has initial velocity,
→
v
=
3
^
i
+
4
^
j
and a constant force
→
F
=
4
^
i
−
3
^
j
acts on it. The path of the particle can be:
Q.
Two particles having position vectors
→
r
=
(
3
→
i
+
5
j
)
m
and
→
r
2
=
(
−
5
i
+
3
j
)
m
are moving with velocities
→
V
1
=
(
4
^
i
−
4
^
j
)
m
s
−
1
and
→
V
2
=
(
a
^
i
−
3
^
j
)
m
s
−
1
. If they collide after
2
seconds , the value of
a
is
Q.
The initial velocity of a particle,
→
u
=
4
→
i
+
3
→
j
. It is moving with uniform acceleration,
→
a
=
0.4
→
i
+
0.3
→
j
. Its velocity after
10
seconds is
Q.
A particle is projected with initial velocity
→
v
=
(
10
^
i
+
15
^
j
)
m
/
s
in
x
−
y
plane. The magnitude of displacement of the particle at time
t
=
1
s
is
(
→
g
=
−
10
^
j
m
/
s
2
)
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