A small body is thrown at an angle to the horizontal with the initial velocity `v_0`. Neglecting the air drag, find:
(a) the displacement of the body as a function of time `r(t)`,
(b) the mean velocity vector `ltlt v gtgt` averaged over the first t seconds and over the total time of motion.

Answer

(a) `r=v_0t+g t^2//2`; (b) `<< v >>_t=v_0+g t//2`, `<< v >>=v_0-g(v_0g)//g^2`.

Explanation

As the body is under gravity of constant accelerating `vecg`, it's velocity vector and displacement vectors are:
`vecv=vec(v_0)+vec(g t)` (1)
and `Deltavecr=vecr=vec(v_0)t+1/2g t^2` (`vecr=0` at `t=0`) (2)
So, `lt vecv gt` over the first t seconds
`lt vecv gt =(Deltavecr)/(Deltat)=(vecr)/(t)=vec(v_0)+(vec(g t))/(2)` (3)
Hence from Eq. (3), `lt vecv gt` over the first t seonds
`lt vecv gt =vecv_0+vecg/2tau` (4)
For evaluating t, take
`vecv*vecv=(vecv_0+vec(g t))*(vecv_0+vec(g t))=v_0^2+2(vecv_0*vecg)t+g^2t^2`
or, `v^2=v_0^2+(vecv_0*vecg)t+g^2t^2`
But we have `v=v_0` at `t=0` and
Also at `t=tau` (Figure). (also from energy conservation)
Hence using this property in Eq. (5)
`v_0^2-v_0^2+2(vecv_0*vecg)tau+g^2tau^2`
As `tau!=0`, so, `tau=-(2(vecv_o*vecg))/(g^2)`
Putting this value of `tau` in Eq. (4), the average velocity over the time of flight
`lt vecv gevecv_0-vec "g"((vecv_0*vecg))/(g^2)`