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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.

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The correct Answer is:
`s=v_(0)t sqrt(2(1-sin theta))`=22 min.
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