Two bodies were thrown simultaneously from the same point, one, straight up, and the other, at an angle of `theta=60^@` to the horizontal. The initial velocity of each body is equal to `v_0=25m//s`. Neglecting the air drag, find the distance between the bodies `t=1.70s` later.

Answer

`l=v_0tsqrt(2(1-sin theta))=22m`

Explanation

The solution of this problem becomes simple in the frame attached with one of the bodies. Let the body thrown straight up be 1 and the other body be 2, then for the body 1 in the frame of 2 from the kinematic equation for constant acceleration:
`vec(r_(12))=vec(r_(0))_((12))+vec(v_0)_((12))t+1/2vecw_(12)t^2`
So, `vec(r_(12))=vec(v_0)_((12))t`, (because `vec(w)_(12)=0` and `vec(r_0)_((12))=0`)
or, `|vec(r_(12))|=|vec(v_0)_((12))|t` (1)
But `|vec(v_(01))|=|vec(v_(02))|=v_0`
So, from properties of triangle
`v_(0(12))=sqrt(v_0^2+v_0^2-2v_0v_0cos(pi//2-theta_0))`
Hence, the sought distance
`|vec(r_(12))|=v_0sqrt(s(1-sintheta)t)=22m`.