35. A body is projected with a velocity 60 ms' at 30° to horizontal. Its initial velocity vector is?
Answers
Answer:Step 1: Initial velocity vector[Refer Fig.]
u
=u
x
i
^
+u
y
j
^
=(ucos30
o
)
i
^
+(usin30
o
)
j
^
=(60cos30
o
)
i
^
+(60sin30
o
)
j
^
u
=(30
3
)
i
^
+(30)
j
^
m/s
→Option (c)
Step 2: Velocity after t= 3 s
In x direction u
x
=30
3
m/s
Velocity in x-direction will not change as there is no acceleration in x direction, therefore v
x
=u
x
In y direction a
y
=−10 m/s
2
;t=3s; u
y
=30 m/s
Since acceleration is constant, therefore applying equation of motion
v
y
=u
y
+at
⇒ v
y
=30−10×3 m/s=0 m/s
V=30
3
i
^
m/s
→Option (d)
Step 3: Displacement after t= 2 s
In x direction: a
x
=0
s
x
=u
x
t+
2
1
a
x
t
2
=30
3
×2
i
^
+0 =60
3
i
^
m
In y direction applying equation of motion
s
y
=u
y
t+
2
1
a
y
t
2
=30×2−
2
1
(10)(2)
2
s
y
=40
j
^
m
∴
s=(60
3
i
^
+40
j
^
)m
→Option (a)
Step 4: Velocity after 2s
In y direction applying equation of motion as the acceleration is constant.
v
y
=u
y
+a
y
t =30−10×2 m/s
=10 m/s
∴
v=(30
3
i
^
+10
j
^
)m/s
→Option (b)
Explanation: