Question

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?

Answer 1

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`