Two bodies were thrown simultaneously from the same point , on straight up and the other, at angle ` theta= 60^@` to the horizontal . The initial velocity of each body is equal to ` u = 30 m//s.` Neglecting the air resistance, find the distance between the bodies after ` 2` seconds.

Explanation

For first body, the distance moved in vetical upward direction in time (t0 is ` y =ut - 1/2 gt^2`
For second body, the horizontal dsitance, Mbrgt ` x' =u cos theta xx t = 30 xx 1/2 = 30 m`
and vertical distance, ` y' = u sin theta t - 1/2 gt^2 `
Vertical distance between two dodies after time (t),
` = y -y' = (ut- 1/2 gt^2) - (u sin theta t- 1/2 gt^2)`
`= ut - u sin theta t`
`= 30 xx 3 - 30 xx (sqrt 3//20 xx 2 ~~ 8m`.
The distance between two boies after tiem (t) is
` = sqrt( (y -y)^2 +x'^2) = sqrt (8^2+30^2) ~~31 m`.