Question

A particle is projected at an angle $\theta ={30}^o$ with the horizontal, with a velocity of $10m{s}^{-1}$. Then

• A. After $2s$, the velocity of particle makes an angle of ${60}^o$ with initial velocity vector.
• B. After $1s$, the velocity of particle makes an angle of ${60}^o$ with initial velocity vector.
• C. The magnitude of velocity of particle after $1s$ is $10m{s}^{-1}$.
• D. The magnitude of velocity of particle after $1s$ is $5m{s}^{-1}$.

Answer 1

Answer: (B), (C)

After $1s$, the velocity of particle makes an angle of ${60}^o$ with initial velocity vector.
The magnitude of velocity of particle after $1s$ is $10m{s}^{-1}$.

$B,C$

$\therefore$ Velocity vector will be,

$\overrightarrow{V}=5\sqrt{3}i+(5-gt)j$

$\overrightarrow{V}=5\sqrt{3}i+(5-10t)j$

Initially, ${\overrightarrow{V}}_i=5\sqrt{3}i+5j$

at $t=1$, ${\overrightarrow{V}}_1=5\sqrt{3}i-5j$

$\therefore$ $\theta ={Cos}^{-1}\frac{{\overrightarrow{V}}_i.{\overrightarrow{V}}_1}{\left|{\overrightarrow{V}}_i\right|\left|{\overrightarrow{V}}_1\right|}$

$={Cos}^{-1}\frac{75-25}{10\times 10}={Cos}^{-1}\left(\frac{50}{100}\right)$

$={60}^0$

$\left|{\overrightarrow{V}}_1\right|=\sqrt{{\left(5\sqrt{3}\right)}^2+{(-5)}^2}=10m/s$