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A particle is projected vertically upwar...

A particle is projected vertically upwards from O with a velocity and a second particle is projected at the same instant from P (at a height h above O) with velocity v at an angle of projection `theta` with the horizontal. Find the time when the distance between them is minimum. Take `v = 5m//s` and `h = 20 m`

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The correct Answer is:
2
In from of ball thrown from p

`v_("resultant") = 2v cos ((90^(@) + theta)/(2))`
`= 2v cos (45^(@) + theta//2)`
`t = (d)/(v_("resultant")) = (h cos (45^(@) + (theta)/(2)))/(2 v cos (45^(@) + (theta)/(2))) = (h)/(2v)`
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