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

For the first body, the distance moved in the vertical upward direction in time t is:

y = ut − ¹/₂gt²

For the second body, the horizontal distance is:

x' = u cosθ · t

and the vertical distance is:

y' = u sinθ · t − ¹/₂gt²

Vertical distance between two bodies after time t:

y − y' = u sinθ · t − ut

Distance between two bodies after time t:

= √[(x')² + (y − y')²]

= √[(u cosθ · t)² + (u sinθ · t − ut)²]

= ut √[(cos²θ + sin²θ) + 1 − 2 sinθ]

= ut √[2(1 − sinθ)]

Substituting u = 30 m/s, t = 2 s, θ = 60°:

= 30 × √2(1 − sin 60°) = 30 m