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    				| 1. | 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)` | 
| Answer» Correct Answer - 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`. | |