A body is thrown with the velocity $20 \mathrm{m} {\mathrm{s}}^{-1}$ at an angle of $60°$ with the horizontal. Find the time gap between the two positions of the body where the velocity of the body makes an angle of $30°$ with horizontal

• Correct answer
  A

  $1.15 \mathrm{s}$

   
• B

  $0.95 \mathrm{s}$
• C

  $1 \mathrm{s}$
• D

  $1.5 \mathrm{s}$

   

Solution

$\tan \alpha =\frac{V_y}{V_x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{{20}^{\text{o}}\sin 60-10t_1}{20.\cos 60}$

$\Rightarrow t_1=\frac{2}{\sqrt{3}}$

$\frac{-1}{\sqrt{3}}=\frac{20\sin 60-10t_2}{20\cos {60}^{\text{o}}}$

$t_2=\frac{4}{\sqrt{3}}$

$\Rightarrow \Delta t=\frac{2}{\sqrt{3}}=1.15\text{ s}$

Hence, the correct answer is option: 'A'.