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    				| 1. | In the figure shown, find. (a) time of flight of the projectile along the inclined plane. (b) range OP. | 
| Answer» Correct Answer - A::B (a) Horizontal component of initial velocity, `u_x = 20sqrt(2) cos 45^@ = 20 m//s` Vertical component of initial velocity, `u_y = 20 sqrt(2) sin 45^@ = 20 m//s` Let the particle, strikes the inclined plane at P after time t, then horizontal displacement `QP = u_x t = 20t` In vertical displacement, `u_yt or 20t ` is upward and `1/2 g t^2 or 5t^2` is downwards. But net vertical displacement is downwards. Hence `5t^2` should be greater than 20 t and therefore, `OQ = 5t^2- 20t` In `DeltaOQP, ` `tan37^@ = OQ/QP` or `3/4 = (5t^2 - 20t)/(20t)` Solving this equation, we get `t= 7s` (b) Range, OP `= (PQ) sec 37^@` `= (20t)(5/4)` Substituting the value of t, we get `OP = 175 m.` | |