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A particle is projected from ground with...

A particle is projected from ground with velocity `20(sqrt2) m//s` at `45^@`. At what time particle is at height 15 m from ground? `(g = 10 m//s^2)`

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The correct Answer is:
A, C, D
Vertical component of initial velocity
`u = 20(sqrt2) sin 45^@ = 20 m//s`
Now, apply `s = ut + 1/2 at^2` (to find t) in vertical
direction, with
`s = 15m, u=20 m//s and a = -10 m//s^2`.
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