A particle is projected from ground with speed 80m/s at an angle 30° with horizontal from ground.What is the magnitude of average velocity of particle in time interval t=2s to t=6s?

1 Answer
Mar 18, 2018

Let's see the time taken by the particle to reach maximum height,it is, t=(u sin theta)/g

Given,u=80ms^-1,theta=30

so,t=4.07 s

That means at 6s it already started moving down.

So,upward displacement in 2s is, s=(u sin theta)*2 -1/2 g (2)^2=60.4m

and displacement in 6s is s=(u sin theta)*6 - 1/2 g (6)^2=63.6m

So,vertical dispacement in (6-2)=4s is (63.6-60.4)=3.2m

And horizontal displacement in (6-2)=4s is (u cos theta*4)=277.13m

So,net displacement is 4s is sqrt(3.2^2 +277.13^2)=277.15m

So,average velcoity = total displacement /total time=277.15/4=69.29 ms^-1