A particle is projected vertically upward with a velocity V₀ ,but due to horizontal wind the particle experiences a horizontal constant accleration of a₀ m/s² . The horizontal distance travelled by the particle in a time interval in which its velocity vector rotates by 90⁰ is_____________.

Dear sindhuja

Initial velocity vector =V₀ j

when particle will be at max hight then it has no verticle velocity,it has velocity in the horizontal direction only.

final velocity vector =V₁ i

and this will be the position where velocity vector rotates by 90 degree.

so time taken to reach highest point is 

0=V₀ -gt

t=V₀/g

and in this time horizontal distance travelled

S=0 + 1/2 *a₀*(V₀/g)²

S=a₀V₀²/2g²

---

Please feel free to post as many doubts on our discussion forum as you can.
 If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin