Physics, asked by akash4563singh, 20 days ago

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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Answered by deepanshu20753
0

Answer:

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Class 11

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>>Motion in a Plane

>>Projectile Motion

>>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.

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