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#