Question

A particle is moving in xy plane so that its position vector varies with time as r is equal to 10 root 3 ICAP + 20 - 30 square if initial velocity of the particle is inclined at an angle theta from the vertical then theta is​

Answer 1

Answer:

The angle made with the vertical is $60^\circ$

Explanation:

The position vector of the particle varies with time as

$\vec r=10\sqrt{3}t\hat i+(10t-t^2)\hat j$

We know that

The velocity vector is given by

$\boxed{\vec v=\frac{d\vec r}{dt}}$

$\implies \vec v=\frac{d}{dt}(10\sqrt{3}t\hat i+(10t-t^2)\hat j$

$\implies \vec v=10\sqrt{3}\frac{dt}{dt}\hat i+\frac{d}{dt}(10t-t^2)\hat j$

$\implies \vec v=10\sqrt{3}\hat i+(10-2t)\hat j$

For initial velocity t=0

Therefore, the initial velocity

$\vec v=10\sqrt{3}\hat i+10\hat j$

The angle made by the initial velocity vector from the horizontal

$\theta=\tan^{-1}\frac{10}{10\sqrt{3}}$

$\theta=\tan^{-1}\frac{1}{\sqrt{3}}$

$\theta=30^\circ$

Therefore, the angle made with the vertical = $90^\circ-30^\circ=60^\circ$

Hope this is helpful.