A projectile is launched with initial velocity vo-121 + 3)) m/s. The equation of trajectory of particle is (g = 10 m/s2
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Class 11
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>>A projectile is launched with an initial
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A projectile is launched with an initial velocity
V
0
=(2m/s)
i
^
+(3m/s)
j
^
. At the top of the trajectory, the speed of the particle is (x horizontal direction, y-vertical dirction)-
Hard
Solution
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Correct option is B)
Given,
v
0
=(2m/s)
i
^
+(3m/s)
j
^
At the top of the trajectory, the vertical component of the speed gets vanished.
So, v
y
=0m/s
And the Horizontal component of the speed,
v
x
=2m/s
At the top of the trajectory, speed of the particle is
v
=v
x
i
^
+v
y
j
^
v
=2
i
^
+0
j
^
v=∣
v
∣=
(2)
2
+(0)
2
v=2m/s
The correct option is B.