Question

A particle is projected with velocity 10m/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:

The vertical component of velocity vector is 5√3/3 m/s in the downward direction

Explanation:

The angle of projection θ = 60°

Initial velocity u = 10 m/s

Resolving the initial velocity into two components

The horizontal component

$u_x=10\cos60^\circ=5\text{ m/s}$

Vertical component

$u_y=10\sin60^\circ=5\sqrt{3}\text{ m/s}$

Therefore the velocity vector can be written as

$\vec u=5\hat i+5\sqrt{3}\hat j$

Throughout the motion of the projectile the horizontal component of the velocity will not change

Let at any instant its velocity vector be

$\vec v=5\hat i+y\hat j$

If this velocity is perpendicular to the initial velocity then the dot product of both will be zero

Therefore

$\vec u.\vec v =0$

$\implies (5\hat i+5\sqrt{3}\hat j).(5\hat i+y\hat j)=0$

$\implies 25+5\sqrt{3}y=0$

$\implies y=-5\sqrt{3}/3$

Therefore the vertical component of velocity vector is 5√3/3 m/s

Answer 2

Answer:

the answer is 5/√3 m/s

as simple as that