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A particle is projected with velocity u ...

A particle is projected with velocity u at angle `theta` with horizontal. Find the time when velocity vector is perpendicular to initial velocity vector.

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You can arrive at the answer of this problem by many different ways, the degree of difficuity depends largely on the choise of reference frame. The various method given below.
Let after time t at point P the velocity v becomes perpendicular to the direction of initial velocity as shown in the figure. The horizontal of projectile remains same at all instants

` Rightarrow v sin theta = u cos theta`
The vertical component after time t.
` - v cos theta = u sin theta - "gt"`
Dividing equation (i) by (ii) we get
` - tan theta - ( u cos theta)/( u sin theta - "gt")`
` Rightarrow -u sin^(2) theta + "gt" sin theta = u cos^(2) theta " " ( " put" tan theta = (sin theta)/(cos theta))`
` Rightarrow gt sin theta = u`
` Rightarrow t = u/( g sin theta) `
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