Question

At high altitude, a body explodes at rest into two equal fragments with one fragment receiving horizontal velocity of $10m/s$. Time taken by the two radius vectors connecting point of explosion to fragments to make ${90}^{\circ }$ is

• A. $10s$
  
• B. $1s$
  
• C. $2s$
  
• D. $4s$
  

Answer 1

The correct option is C $2s$


Draw a diagram of given situation.

Find the time taken by the particle to make ${90}^{\circ }$.
Formula used: $s=ut+\frac{1}{2}a{t}^2$
According to given situation,
$\theta +\theta ={90}^{\circ }\Rightarrow \theta ={45}^{\circ }$
Horizontal velocity $=10m/s$
Horizontal distance covered by particle,
$s=ut+\frac{1}{2}a{t}^2$
$x=10t$
Vertical distance covered by particle,
$s=ut+\frac{1}{2}a{t}^2$
$y=\frac{1}{2}g{t}^2$
From above equations,
$\tan \theta =\frac{x}{y}=\frac{10t}{\frac{1}{2}g{t}^2}=\frac{2}{t}$
$tan{45}^{\circ }=\frac{2}{t}\Rightarrow t=2s$
Final Answer: $\left(C\right)$