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 Two objects are located at a height 10 m above the ground. At some point of time, the objects are thrown with initial velocity $2\sqrt{2}$ m/s at an angle $45^{0}$ and $135^{0}$ with the positive X-axis, respectively. assuming  g=10m/s² , the velocity vectors will be perpendicular to each other at a time is equal to 

Answer:

Explanation:

 In this projectile motion, both objects are projected at the angle difference of $90^{0}$  . They will again have same angle  difference when they are again at same height (of 10 m) from ground. Taking , plane $O_{2}OO_{1}$  as reference , time taken to reach  at point $O_{1}$  by first object is

 Time of flight  . $T_{1}= \frac{2u \sin \theta}{g}$

   $=\frac{2\times2\sqrt{2} \sin 45^{0}}{10}=0.4 s$

Time taken  by second object to reach at point $O_{2}$ is 

  $T_{2}=\frac{2\times2\sqrt{2} \sin 135^{0}}{10}=0.4 s$

 $\because$    $T_{1}=T_{2}$

 So, velocities of both objects will be perpendicular  to each other after 0.4 s