Question

A particle is projected with velocity 10 m/s at an
angle of 60° from ground. Then the vertical
component of velocity vector when instantaneous
velocity becomes perpendicular to initial velocity, is​

Answer 1

Answer:

$Vertical\ component\ of\ velocity=\dfrac{5}{\sqrt 3}$

Explanation:

Given that  

Velocity V =10 m/s

 θ=60°

Lets take velocity of particle is U when it will become perpendicular to the initial velocity V.

 At initial condition

The horizontal component of velocity = 10 cos 60°

At final condition  horizontal component of velocity = U sin 60°

And we know that in projectile motion horizontal component of velocity will remain constant .So

10 cos 60°=U sin 60°

U=10 cot 60°

$U=\dfrac{10}{\sqrt{3}}$

Now the vertical component of velocity =U cos 60°

By putting the value of U

$Vertical\ component\ of\ velocity=\dfrac{10}{\sqrt 3}\times \dfrac{1}{2}$

$Vertical\ component\ of\ velocity=\dfrac{5}{\sqrt 3}$